# HBSE 9th Class Maths Important Questions Chapter 7 Triangles

Haryana State Board HBSE 9th Class Maths Important Questions Chapter 7 Triangles Important Questions and Answers.

## Haryana Board 9th Class Maths Important Questions Chapter 7 Triangles

Question 1.
In ΔPQR, if altitude PM bisects QR, prove that PQ = PR.
Solution :
Given: A ΔPQR such that PM ⊥ QR and MQ = MR.
To prove PQ = PR.

Proof: In ΔPMQ and ΔPMR, we have
MQ = MR, (Given)
∠PMQ = ∠PMR, (Each = 90°)
and PM = PM, (Common)
∴ ΔPMQ ≅ ΔPMR,
(By SAS congruence rule)
⇒ PQ = PR (CPCT)
Hence Proved

Question 2.
In the figure, BD ⊥ AC, CE ⊥ AB and AB = AC. Prove that BD = CE.

Solution :
In ΔABD and ΔACE, we have
∠ADB = ∠AEC (Each = 90°)
and AB = AC (Given)
∴ ΔABD ≅ ΔACE
(By AAS congruence rule)
⇒ BD = CE (CPCT)
Hence Proved

Question 3.
In the figure, ABCD is a quadrilateral in which AX ⊥ BD, CY ⊥ BD, AX = CY and BX = YD. Show that :
(i) ΔAXD ≅ ΔCYB

Solution :
(i) We have
BX = YD
⇒ BX + XY = YD + XY,
⇒ BY = XD ……(i)
Now in ΔAXD and ΔCYB, we have
AX = CY. (Given)
∠AXD = ∠CYB (Each = 90°)
and XD = BY, [From (i)]
∴ ΔAXD ≅ ΔCYB, (By SAS congruence rule)
Hence proved

(ii) ΔАХD ≅ ΔCYB (As proved above)
Hence proved

Question 4.
On the arms AB and BC of an ∠ABC, points N and M are taken respectively such that ∠MAB = ∠NCB (see the figure). If AB = BC, then prove that BM = BN.

Solution :
In ΔABM and ΔCBN, we have
∠MAB = ∠NCB, (Given)
AB = BC, (Given)
∠B = ∠B (Common)
∴ ΔABM ≅ ΔCBN,
(By ASA congruence rule)
⇒ BM = BN, (CPCT)
Hence proved

Question 5.
In the figure, AP ⊥ BD, CQ ⊥ BD and AP = CQ. Prove that BD bisects AC.

Solution :
In ΔAOP and ΔCOQ, we have
∠APO = ∠CQO,
[∵ AP ⊥ BD and CQ ⊥ BD
∴ Each = 90°]
∠AOP = ∠COQ,
(Vertically opposite angles) and
and AP = CQ, (Given)
∴ ΔAOP ≅ ΔCOQ,
(By AAS congruence rule)
⇒ AO = OC, (CPCT)
⇒ O is the midpoint of AC.
Hence, BD bisects AC. Hence Proved

Question 6.
See figure, explain how one can find the breadth of the river without crossing it.

Solution :
AB be the breadth of river Mark Point P on the bank of river.
Construction: Let O be the midpoint of BP. Mark a point Q on AO produced such that AO = OQ. Join PQ.
Now in ΔAOB and ΔQOP, we have
BO = OP
(O is the midpoint of BP)
∠AOB = ∠QOP
(Vertically opposite angles)
and AO = OQ (By construction)
∴ ΔAOB ≅ ΔQOP
(By SAS congruence rule)
⇒ AB = PQ (CPCT)
Hence, breadth of river AB = PQ i.e., one should measure PQ to find the breadth AB of the river.

Question 7.
In the figure, AP is the bisector of ∠CAD and AP || BC. Prove that AB = AC.

Solution :
∠DAP = ∠CAP …………..(i)
(∵ AP is the bisector of ∠CAD)
∵ AP || ВC (Given)
∠DAP = ∠ABC …………..(ii)
(Corresponding angles)
∠CAP = ∠ACB ………….(iii)
(Alternate interior angles)
From (i), (ii) and (iii), we get
∠ABC = ∠ACB
⇒ AB = AC, (sides opposite to equal angles are equal)
Hence proved

Question 8.
In the figure, AC = BC and ∠x = ∠y.
Prove that: (i) ΔABD ≅ ΔBAE (ii) AD = BE.

Solution :
(i) We have, BC = AC, (Given)
⇒ ∠B = ∠A, …………(i)
(Angles opposite to equal sides are equal)
In ΔABD and ΔBAE, we have
∠B = ∠A [from (i)]
∠y = ∠x, (Given)
and AB = AB, (Common)
∴ ΔABD ≅ ΔBAE, (By AAS congruence rule)
Hence proved

(ii) ΔABD ≅ ΔBAE (CPCT)

Question 9.
ABCD is a quadrilateral in which AB = BC and AD = CD. Show that BD bisects both the angles ABC and ADC.
[NCERT Exemplar Problems]

Solution :
In ΔABD and ΔCBD, we have
AB = BC (given)
and BD = BD (common)
∴ ΔABD = ΔCBD
(by SSS congruence rule)
⇒ ∠ABD = ∠CBD (CPCT) and
So, BD bisects ∠ABC and ∠ADC.
Hence proved

Question 10.
In a triangle ABC, if ∠A = 55° and ∠C = 65°. Determine the shortest and longest sides of the triangle.
Solution :
In a ΔABC, we have
∠A = 55°, ∠C = 65°
But
∠A + ∠B + ∠C = 180°
(Sum of interior angles of a triangle = 180°)

⇒ 55° + ∠B + 65° = 180°
⇒ 120° + ∠B = 180°
⇒ ∠B = 180° – 120°
⇒ ∠B = 60°
∵∠C is the greatest angle.
∴ AB is the longest side of the ΔABC. [∵ Side opposite to greater angle is larger] and ∠A is the smallest angle.
∴ BC is the shortest side of the ΔABC. [∵ side opposite to smaller angle is shortest]
Hence, AB is the longest side and BC is the shortest side of the triangle.

Question 11.
In figure, D is a point on the side AC of ΔABC and E is a point such that CD = ED. Prove that AB + BC > AE.

Solution :
In ΔABC, we have
AB + BC > AC, (Sum of any two sides of a triangle is greater than third side)
⇒ AB + BC > CD + AD,
[∵ AC = CD + AD]
⇒ AB + BC > ED + AD ……(i)
[∵ It is given that CD = ED]
In ΔAED, we have
AD + ED > AE ……(ii)
From (i) and (ii), we get
AB + BC > AE. Hence proved

Question 12.
The ABCD is a rectangle in which sides AB and AD produced to E and respectively such that AB = BE and EC = CF. Prove that AD = DF.
Solution :
We have,
AB = BE, (Given)
AB = DC, (Opposite sides of rectangle)
∴ BE = DC ……(i)

In ΔEBC and ΔCDF, we have
∠EBC = ∠FDC, (Each = 90°)
Hyp. EC = Hyp. CF, (Given)
and BE = DC,
[As proved above in (i)]
∴ ΔEBC ≅ ΔCDF
(By RHS congruence rule)
⇒ BC = DF, (CPCT) … (ii)
From (ii) and (iii), we get

Question 13.
In figure, ABCD is a square. P and Q are points on sides AB and CD respectively such that CP = BQ. Prove that :
(i) BP = QC
(ii) ∠BCP = ∠CBQ

Solution :
In ΔPBC and ΔQCB, we have
∠PBC = ∠QCB,
(Each angle of a square is 90°)
Hyp. PC = Hyp. BQ, (Given)
and BC = BC (Common)
∴ ΔPBC ≅ ΔQCB,
(By RHS congruence rule)
⇒ PB = QC, (CPCT) [Proved (i)]
and ∠BCP = ∠CBQ,(CPCT) [Hence proved (ii)]

Question 14.
In the figure, ABCD is a quadrilateral in which AD = BC. If equal perpendiculars DP and BQ are drawn on diagonal AC. Prove that AP = CQ

Solution :
In ΔDPA and ΔBQC, we have
∠DPA = ∠BQC,
[∵ DP ⊥ AC and BQ ⊥ AC]
Hyp. AD = Hyp. BC, (Given)
and DP = BQ (Given)
∴ ΔDPA ≅ ΔBQC,
(By RHS congruence rule)
⇒ AP = CQ, (CPCT)
Hence proved

Question 1.
In the ΔABC, D is the midpoint of BC, AD is produced upto E so that DE = AD.
Prove that:
(i) ΔABD ≅ ΔECD
(ii) AB = EC
(iii) AB || EC
Solution :
Given: A ΔABC, in which D is the midpoint of BC and AD is produced upto E such that AD = DE.
To prove : (i) ΔΑΒD ≅ ΔECD
(ii) AB = EC
(iii) AB || EC

Proof :
(i) In ΔΑΒD and ΔECD, we have
BD = CD
(∵ D is the midpoint of BC)
∠ADB = ∠EDC, (vertically opposite angles)
∴ ΔABD ≅ ΔECD
(by SAS congruence rule) Hence proved

(ii) ΔABD ≅ ΔECD
(As proved above)
⇒ AB = EC, (CPCT) Hence proved

(iii) ΔABD ≅ ΔECD
(As proved above)
⇒ ∠ABD = ∠ECD, (CPCT)
⇒ ∠ABC = ∠ECB,
But these are alternate interior angles.
So, AB || EC, (By theorem 6.3) Hence proved

Question 2.
In the figure, ABCD is a parallelogram in which E is the mid point of BC. DE is produce and intersect side AB produced at L. Prove that AL = 2CD.

Solution :
Since opposite sides of a parallelogram are equal and parallel.
∴ AB = CD
and AB || CD …………(i)
∵ AB || CD and DL is the transversal.
∴ ∠CDL = ∠ALD
(A pair of alternate interior angles)
⇒ ∠CDL = ∠BLE …………(ii)
Now in ΔCDE and ΔBLE, we have
∠CDE = ∠BLE, [from (ii)]
∠DEC = ∠LEB,
(vertically opposite angles)
and EC = EB,
(E is the midpoint of BC)
∴ ΔCDE ≅ ΔBLE,
(By AAS congruence rule)
⇒ CD = BL (CPCT)
⇒ AB = BL, [Using (i) …(iii)]
Now, AL = AB + BL
⇒ AL = AB + AB, [Using (iii)]
⇒ AL = 2AB
⇒ AL = 2CD, [Using (i)]
Hence proved

Question 3.
In the parallellogram ABCD, the angles A and Care obtuse. Points P and Q are taken on the diagonal BD such that the angles PAD and QCB are right angles. Prove that PA = QC.
Solution :
Given: A parallelogram ABCD such that ∠A and ∠C are obtuse and ∠PAD = ∠QCB = 90°.
To prove: PA = QC.

Proof : Since opposite sides of a parallelogram are parallel and equal.
∵ AD || BC and BD is the transversal
⇒ ∠ADB = ∠CBD, (A pair of alternate interior angles)
Now, in ΔPAD and ΔQCB, we have
∠PAD = ∠QCB, (Each = 90°)
and ∠ADP = ∠CBQ, [From (ii)]
(By ASA congruence rule)
⇒ PA = QC, (CPCT)
Hence proved

Question 4.
In the figure, ABCD is a square in which P, Q and R are the points in AB, BC and CD respectively such that AP = BQ = CR. Prove that:
(i) PB = QC
(ii) PQ = QR

Solution :
∵ ABCD is a square and we know that each side of a square is equal.
∴ AB = BC = CD = DA
Now, AB = BC
⇒ AB – AP = BC – AP
(Subtracting AP from both sides)
⇒ AB – AP = BC – BQ (∵ AP = BQ)
⇒ PB = QC …………..(i) Hence Proved.

(ii) Now in triangles PBQ and QCR, we have
PB = QC [From (i)]
∠PBQ = ∠QCR
[∵ Each angle of a square is 90°]
and BQ = CR (Given)
ΔPBQ ≅ ΔQCR
(By SAS congruence rule)
⇒ PQ = QR (CPCT) Hence Proved.

Question 5.
In the figure, ΔABC is the right-angled at B. Squares ABPQ and ACDE are draw on the sides AB and AC of ΔABC respectively. Prove that :
(i) ΔQAC ≅ ΔBAE
(ii) QC = BE

Solution :
(i) We have
∠QAB = ∠CAE
(Each angle of the square is 90°)
⇒ ∠QAB + ∠BAC = ∠CAE + ∠BAC,
⇒ ∠QAC = ∠BAE ……(i)
Now, in ΔQAC and ΔBAE, we have
AQ = AB, (Equal sides of square ABPQ)
∠QAC = ∠BAE, [From (i)]
and AC = AE, (Equal sides of square ACDE)
∴ ΔQAC ≅ ΔBAE, (By SAS congruence rule)
Hence proved

(ii) ΔQAC ≅ ΔBAE,
(As proved above)
⇒ QC = BE (CPCT)
Hence proved

Question 6.
In figure, the lines segment joining the midpoints P and Q of opposite sides AD and BC of quadrilateral ABCD is perpendicular to both these sides. Prove that the other sides of the quadrilateral are equal. [NCERT Exemplar Problems]

Solution :
Join BP and CP.
In ΔPQB and ΔPQC, we have
BQ = QC,
[∵ Q is the midpoint of BC]
∠PQB = ∠PQC, [Each = 90°]
and PQ = PQ, (Common)
∴ ΔPQB ≅ ΔPQC,
(By SAS congruence rule)
⇒ BP = CP, (CPCT) …(i)
and ∠BPQ = ∠CPQ, (CPCT) …(ii)
Now, ∠APQ = ∠DPQ, (Each = 90°)
⇒ ∠APQ – ∠BPQ = ∠DPQ – ∠BPQ,
(Subtracting ∠BPQ from both sides)
⇒ ∠APQ – ∠BPQ = ∠DPQ – ∠CPQ,
[Using (ii)]
⇒ ∠APB = ∠DPC …(iii)
Now, in ΔAPB and ΔDPV, we have
AP = PD, (∴ Pis the midpoint of AD)
∠APB = ∠DPC, [From …(iii)]
and BP = CP, [From (i)]
∴ ΔAPB ≅ ΔDPC,
(By SAS congruence rule)
⇒ AB = CD, (CPCT)
Or other sides of the quadrilateral are equal. Hence proved

Question 7.
If the diagonals of a quadrilateral bisect each other at right angles and ∠A = 90° prove that ABCD is a square.
Solution :
Given: A quadrilateral ABCD, in which diagonals AC and BD bisect each other at 90° and ∠A = 90°
To prove: ABCD is a square
i.e., AB = BC = CD = DA and
∠A = 90°.

Proof: In ΔAOD and ΔAOB, we have
OD = OB (Given, diagonals bisect each other)
∠AOD = ∠AOB, (Each = 90°)
and AO = AO (Common)
∴ ΔAOD ≅ ΔAOB
(By SAS congruence rule)
⇒ AD = AB (CPCT) …(i)
Similarly we can prove that
ΔAOD ≅ ΔAOB
(By SAS congruence rule)
⇒ AB = BC ……(ii)
(CPCT)
and ΔBOC ≅ ΔDOC
(By SAS congruence rule)
⇒ BC = CD …….(iii) (CPCT)
From (i), (ii) and (iii), we get
AD = AB = BC = CD
and ∠A = 90 (Given)
Hence, ABCD is a square. Proved

Question 8.
In the figure, ∠ABC = 70° and AB = AC = CD. Find the value of x.

Solution :
In the ΔABC, we have AB = AC, (Given)
⇒ ∠ABC = ∠ACB, [∵ Angles opposite to equal sides are equal]
⇒ 70° = ∠ACB [∵ ∠ABC = 70°]
⇒ ∠ACB = 70°
Now in ΔACD, we have
AC = CD (Given)
[∵ Angles opposite to equal sides are equal]
In ΔACD, we have
[By theorem 6.8]
⇒ ∠ADC = $$\frac {70°}{2}$$ = 35° ……….(ii)
(Now in ΔABD, we have
⇒ x° = ∠ABC + ∠ADC
⇒ x° = 70° + 35°,
[∵∠ABC = 70° and from (ii), ∠ADC = 35°]
⇒ x° = 105°
Hence, x° = 105°

Question 9.
In the figure, ABC is an isosceles triangle with AB = AC. BE and CF are respectively bisectors of ∠B and ∠C. Prove that ΔFCB ≅ ΔEBC.

Solution :
In ΔABC, we have AB = AC (Given)
⇒ ∠ABC = ∠ACB ……….(i)
(Angles opposite to equal sides are equal)
⇒ $$\frac {1}{2}$$∠ABC = $$\frac {1}{2}$$∠ACB (Multiply by 1/2 on both sides)
⇒ ∠EBC = ∠FCB ……..(ii)
[∵ BE and CF are the bisectors of ∠B and ∠C respectively]
Now in ΔFCB and ΔEBC,
∠FBC = ∠ECB, [From (i)]
∠FCB = ∠EBC, [From (ii)]
and BC = BC, (Common)
∴ ΔFCB ≅ ΔEBC, (By ASA congruence rule)

Question 10.
If bisector of the vertical angle of a triangle bisects the base, show that the triangle is isosceles. [NCERT Exemplar Problems]
Solution :
Given: A ΔABC in which AP is the bisector of vertical angle A bisects BC on Pie., BP = CP.
To prove AB = AC.

Construction: Produce AP to Q, so that AP = PQ. Join C to Q.
Proof: In ΔABP and ΔQCP, we have
BP = CP, (∵ AP bisects BC)
∠APB = ∠QPC,
(Vertically opposite angles)
and AP = PQ. (By construction)
∴ ΔABP ≅ ΔQCP,
(By SAS congruence rule)
⇒ AB=QC, (CPCT) …(i)
and ∠1 = ∠3, (CPCT) …(ii)
But ∠1 = ∠2, (Given) …(iii)
From (ii) and (iii), we get
∠2 = ∠3
⇒ AC = QC, …(iv)
(Sides opposite to equal angles are equal)
From (i) and (iv), we get
AB = AC
So, ΔABC is an isosceles triangle. Hence proved

Question 11.
In the figure, equilateral triangles APB and AQC are drawn on the sides of a ΔABC. If AB = AC, prove that CP = BQ.

Solution :
Since, ΔAPB and ΔAQC are equilateral triangles.
AB = PB …(i) [Sides of equilateral triangle)
and AC = CQ ………(ii)
AB = AC (Given)…(iii)
From (i), (ii) and (iii), we get
PB = CQ …(iv)
∠ABP = ∠ACQ, ………….(v)
[Each = 60°]
∠ABC = ∠ACB, …(vi)
[∵ AB = AC]
Adding (v) and (vi), we get
∠ABP + ∠ABC = ∠ACQ + ∠ACB
⇒ ∠PBC = ∠QCB …(vii)
Now in ΔPBC and ΔQCB, we have
PB = CQ, [From (iv)]
∠PBC = ∠QCB, (From (vii)]
and BC = BC (Common)
∴ ΔPBC ≅ ΔQCB,
(By SAS congruence rule)
⇒ CP = BQ. (CPCT)
Hence proved

Question 12.
In the figure, CP is the bisector of ∠C meets AB on P. A point Q lies on CP such that AP = AQ. Prove that ∠CAQ = ∠ABC.

Solution :
In ΔAPQ, we have
AP= AQ
∠APQ = ∠AQP ……….(i)
[Angles opposite to equal sides are equal]
∵ CP is the bisector of ∠C.
⇒ ∠ACP = ∠BCP ……….(ii)
In ΔBPC, we have
∠APC = B + ∠BCP,
[By theorem 6.8]
⇒ ∠APQ = ∠B + ∠ACP, ……(iii)
[Using (ii)]
In ΔAQC, we have
∠AQP = ∠CAQ + ∠ACQ
[By theorem 6.8]
⇒ ∠AQP = ∠CAQ + ∠ACP …(iv)
From (i), (iii) and (iv), we get
∠B + ∠ACP = ∠CAQ + ∠ACP
⇒ ∠B = ∠CAQ
⇒ ∠ABC = ∠CAQ. Hence proved

Question 13.
In a parallelogram ABCD diagonals AC and BD are equal. Find the measure of angle B.

Solution :
We have, ABCD is a parallelogram. AC and BD are its diagonals.
∴ AB = CD, ……(i) (Opposite sides of a parallelogram)
Now in ΔABC and ΔDCB, we have
AB = CD, [From (i)]
AC = BD, (Given)
and BC = BC, (Common)
∴ ΔABC ≅ ΔDCB.
(By SSS congruence rule)
⇒ ∠ABC = ∠DCB, (CPCT) …(ii)
Now AB || CD and transversal BC intersects them at B and C respectively. ∠ABC + ∠DCB = 180°
(Sum of co-interior angles is 180°)
⇒ ∠ABC + ∠ABC = 180°
[From (ii), ∠DCB = ∠ABC]
⇒ 2∠ABC = 180°
⇒ ∠ABC = $$\frac {180°}{2}$$ = 90°
Hence measure of ∠B = 90°

Question 14.
In the figure, ΔPQR is a triangle in which PM ⊥ QR. Prove that :
(i) PQ > PM
(ii) PQ + PR > 2PM.

Solution :
(i) In right-angled ΔPQM, we have
Proof : ∠PQM + ∠QPM + ∠PMQ = 180°,
(Sum of interior angles of a triangle = 180°)
⇒ ∠PQM + ∠QPM + 90° = 180°
⇒ ∠PQM + ∠QPM = 180° – 90°
⇒ ∠PQM + ∠QPM = 90°
∴ ∠PQM and ∠QPM are acute angles.
⇒ ∠PMQ > ∠PQM
⇒ PQ > PM ………….(i)
(Side opposite to greater angle is larger)
Hence proved

(ii) Similarly, In right-angled ΔPRM, we have
∠PRM + ∠RPM = 90°
⇒ ∠PRM and ∠RPM are acute angles.
⇒ ∠PMR > ∠PRM
⇒ PR > PM ………….(ii)
(Side opposite to greater angle is larger)
Adding (i) and (ii), we get
PQ + PR > PM + PM
⇒ PQ + PR > 2PM. Hence proved

Question 15.
In the figure, BM and CM are the bisectors of ∠B and ∠C respectively and AC > AB. Prove that CM > BM.

Solution :
In ΔABC, we have
AC > AB (Given)
⇒ ∠B > ∠C [Angle opposite to longer side is greater]
⇒ $$\frac {1}{2}$$∠B > $$\frac {1}{2}$$∠C
(Multiply by 1/2 on both sides)
⇒ ∠MBC > ∠MCB, [∵ BM and CM are bisectors of ∠B and ∠C respectively]
∴ ∠MBC = $$\frac {1}{2}$$∠B and ∠MCB = $$\frac {1}{2}$$∠C)
⇒ CM > BM, (Side opposite to larger angle is longer)
Hence proved

Question 1.
In a right-angled triangle, if one of the acute angles is double the other, prove that the hypotenuse is double the shortest side. [NCERT Exemplar Problems]
Solution :
Given: A ΔPQR such that
∠PQR = 90° and ∠R = 2.∠P.
To prove: PR = 2RQ.
Construction: Produce RQ to S such that QS = QR. Join PS.
Proof: ∠R = 2∠P, (Given)
Let ∠P = x, ∠R = 2x
In right-angled ΔPQR, we have
∠P + ∠R + ∠Q = 180°
(∵ Sum of interior angles of a Δ = 180°)
⇒ x + 2x + 90° = 180°
⇒ 3x = 180° – 90°
⇒ 3x = 90°
⇒ x = $$\frac {90°}{3}$$ = 30°
∴ ∠P = 30° and ∠R = 2 × 30° = 60°
Now, ΔPQR and ΔPQS, we have
QR = QS, (By construction)
∠PQR = ∠PQS, (Each = 90°)
and PQ = PQ, (Common)
∴ ΔPQR ≅ ΔPQS
(By SAS congruence rule)
⇒ ∠R = ∠S (CPCT)
⇒ ∠R = ∠S = 60°
⇒ ΔPRS is an equilateral triangle.
RS = PS = PR, …(i) (Sides of an equilateral triangle)
But RQ = QS, (By construction)
∴ RS = RQ + QS
⇒ RS = RQ + RQ
⇒ RS = 2RQ
⇒ PR = 2RQ, [Using (i)]
Hence proved

Question 2.
Prove that the medians of an equilateral triangle are equal.
Solution :
Given: An equilateral ΔABC such that AL, BM and CN are its medians.
To prove: AL = BM = CN.

Proof: ΔABC is an equilateral triangle.
∴ AB = BC = AC ……………..(i)
⇒ AB = BC
⇒ $$\frac {1}{2}$$AB = $$\frac {1}{2}$$BC
(Multiply by 1/2 on both sides)
⇒ AN = BL …………….(ii)
(∵ CN and AL are medians)
Now in ΔALB and ΔCNA, we have
AB = AC, [From (i)]
∠ABL = ∠CAN, (Each = 60°)
and BL = AN, [From (ii)]
∴ ΔALB ≅ ΔCNA,
(By SAS congruence rule)
⇒ AL = CN, (CPCT) …(iii)
Similarly, ΔΑLC ≅ ΔΒΜΑ,
(By SAS congruence rule)
⇒ AL = BM (CPCT) …(iv)
From (iii) and (iv), we get
AL = BM = CN
Hence, medians of an equilateral triangle are equal.
Proved

Question 3.
In the figure, BCDE is a square and ΔABC is an equilateral triangle. Prove that :
(i) ∠BAE = 15°

Solution :
∠ABE = ∠ABC + ∠CBE
∠ABE = 60° + 90°,
⇒ ∠ABE = 150° …….(i)
Similarly, ∠ACD = 60° + 90° = 150° …(ii)
In ΔABE and ΔACD, we have
AB = AC, (Sides of an equilateral triangle)
∠ABE = ∠ACD,
[From (i) and (ii)]
and BE = CD, (Square’s sides)
∴ ΔABE ≅ ΔACD,
(By SAS congruence rule)
Proved (ii)
∵ AB = BC,
(Sides of an equilateral triangle)
and BE = BC, (Square’s sides)
⇒ AB = BE
⇒ ∠BAE = ∠BEA, ………….(iii)
(Angles opposite to equal sides are equal)
Now in ΔABE, we have
∠ABE + ∠BAE + ∠BEA= 180°,
(Sum of interior angles of a triangle = 180°)
⇒ 150° + ∠BAE + ∠BAE = 180°,
[From (iii), ∠BEA = ∠BAE]
⇒ 2∠BAE = 180° – 150°
⇒ ∠BAE = $$\frac {30°}{2}$$
⇒ ∠BAE = 15°. Proved (i).

Question 4.
In ΔPQR, ∠Q = 2∠R, PM is the bisector of ∠QPR meets QR on M and PQ = MR (see figure). Find the ∠PQR.
Solution :
In ΔPQR, we have
∠Q = 2∠R (Given)
Let ∠Q = 2∠R = 2x
⇒ ∠R = x

PM is the bisector of ∠QPR (given).
Let ∠QPM = ∠RPM = y
Draw QN, bisector of ∠PQR.
∴ ∠PQN = ∠RQN = x
In ΔNQR, ∠NQR = ∠QRN = x
⇒ QN = NR …(i) (∵ Sides opposite to equal angles are equal)
In ΔPQN and ΔMRN, we have
PQ = MR. (Given)
∠PQN = ∠MRN = x
and QN = RN, [From (i)]
∴ ΔPQN ≅ ΔMRN,
(By SAS congruence rule)
⇒ ∠QPN = ∠RMN, (CPCT)
⇒ ∠QPN = ∠RMN = 2y …(ii) (CPCT)
and PN = MN, (CPCT)
⇒ ∠NMP = ∠MPN = y, …………(iii)
(Angles opposite to equal sides are equal)
In ΔPQM, we have
∠PMR = ∠PQM + ∠QPM,
(By theorem 6.8)
⇒ ∠PMR = 2x + y ……(iv)
But ∠PMR = ∠NMP + ∠RMN,
⇒ ∠PMR = y + 2y
[Using (ii) and (iii)]
⇒ ∠PMR = 3y ……(v)
From (iv) and (v), we get
2x + y = 3y
⇒ 2x = 2y
⇒ x = y
In ΔPQR, we have
∠P + ∠Q + ∠R = 180°
⇒ 2y + 2x + x = 180°
⇒ 2x + 2x + x = 180°, (∵ y = x)
⇒ 5x = 180°
⇒ x = $$\frac {180°}{5}$$ = 36°
∴ ∠PQR = 2 × 36° = 72°
Hence, ∠PQR = 72°

Question 5.
In the figure, ABCD is a square. X, Y and Z are the points in AB, BC and CD respectively, such that AX = BY = CZ.
Prove that :
(i) XY = YZ
(ii) ∠XYZ = 90°.

Solution :
(i) ∵ ABCD is a square.
∴ AB = BC = CD = DA
Now AB = BC …(i)
and AX = BY (Given)…(ii)
AB – AX = BC – BY
[Subtracting (ii) from (i)]
⇒ BX = CY …….(iii)
(∵ AX = BY)
and BY = CZ (Given) …….(iv)
Now in ΔXBY and ΔYCZ, we have
BX = YC, [From (iii)]
∠XBY = ∠YCZ [Each = 90°]
and BY = CZ, [From (iv)]
∴ ΔXBY = ΔYCZ
(By SAS congruence rule)
⇒ XY = YZ (CPCT)
Hence proved

(ii) ∵ XY = YZ, (As proved above)
⇒ ∠YXZ = ∠YZX [Angles opposite to equal sides one equal]
∴ ∠YXZ = ∠YZX = 45°
In ΔXYZ, we have
∠YXZ + ∠YZX + ∠XYZ = 180°
⇒ 45° + 45° + ∠XYZ = 180°
⇒ 90° + ∠XYZ = 180°
⇒ ∠XYZ = 180° – 90°
⇒ ∠XYZ = 90°. Hence proved

Question 6.
In the figure, ABC is an isosceles triangle in which AB = AC, AE ⊥ BC and F is the mid point of BE and D is the mid
point of EC. Prove that :
(i) E is the midpoint of BC.
(iii) ΔABF ≅ ΔACD.

Solution :
(i) In ΔAEB and ΔAEC, we have
∠AEB = ∠AEC, (Each = 90°)
Hyp. AB = Hyp. AC, (Given)
and AE = AE (Common)
∴ ΔAEB ≅ ΔAEC,
(By RHS congruence rule)
⇒ BE = EC, (CPCT) …(i)
Or E is the midpoint of BC. Hence Proved

(ii) BE = EC, (As proved above)
⇒ $$\frac {1}{2}$$BE = $$\frac {1}{2}$$EC
(Multiply by 1/2 on both sides)
⇒ FE = ED,
[∵ F is the midpoint of BE and D is the midpoint of EC
∴ FE = $$\frac {1}{2}$$BE and ED = $$\frac {1}{2}$$EC]
Now in ΔAFE and ΔADE, we have
FE = ED, (As proved above)
∠AEF = ∠AED, (Each = 90°)
and AE = AE (Common)
(By SAS congruence rule)
⇒ AF = AD, (CPCT) …(ii)
Hence proved

(iii) In ΔABF and ΔACD, we have
AB = AC, (Given)
BF = CD.
[∵ BE = EC ⇒ $$\frac {1}{2}$$BE = $$\frac {1}{2}$$EC
F and D are midpoints of BE and EC respectively ∴ BF = CD]
and AF = AD, [From (ii)]
∴ ΔABF ≅ ΔACD, (By SSS congruence rule)
Hence proved

Question 7.
Show at the difference of any two sides of a triangle is less than the third side.
Solution :
Given: A ΔABC.
To prove :
(i) AC – AB < BC.
(ii) BC – AC < AB.
(iii) BC – AB < AC.

Construction: Take a point D on AC such that AD = AB. Join BD.
Proof: In ΔABD, side AD has been produced to C.
∠BDC > ∠ABD ………..(i)
(∵ Exterior angle of a triangle is greater than each of interior opposite angle)
In ΔBCD, side CD has been produced to A.
[∵ Exterior angle of a triangle is greater than each of interior opposite angle]
In ΔABD, we have
(Angles opposite to equal sides are equal)
From (ii) and (iii), we get
∠ABD > ∠DBC
From (i) and (iv), we get
∠BDC > ∠DBC
⇒ BC > CD, (Side opposite to greater angle is larger)
⇒ CD < BC
⇒ AC – AD < BC
⇒ AC – AB < BC (By construction)
Similarly, BC – AC < AB and BC – AB < AC.
Hence proved

Question 8.
In the figure, ABCD is a square and P, Q and R are points on the sides AB, BC and CD respectively such that AP = BQ = CR. Prove that ∠PQR = 90°.

Solution :
In square ABCD, we have
AP = BQ = CR(Given) …(i)
AB = BC, (Square’s sides)
⇒ AB – AP = BC – AP
(Subtracting AP from both sides)
⇒ AB – AP = BC – BQ,
[From (i), AP = BQ]
⇒ PB = QC ………(ii)
Join PR.
Now, in ΔPBQ and ΔQCR, we have
PB = QC,
[As proved above in (ii)]
∠PBQ = ∠QCR,
(Each angle of square = 90°)
and BQ = CR (Given)
∴ ΔPBQ ≅ ΔQCR
(By SAS congruence rule)
⇒ ∠2 = ∠5, (CPCT) … (iii)
and ∠6 = ∠1 (CPCT)
AB || CD
⇒ PB || CR and PR is the transversal.
⇒ ∠BPR + ∠CRP = 180°,
(Sum of allied angles = 180°)
⇒ ∠2 + ∠3 + ∠1 + ∠4 = 180° …… (iv)
and ∠5 + ∠PQR + ∠6 = 180°, …… (v)
(Linear pair)
From (iv) and (v), we get
∠2 + ∠3 + ∠1 + ∠4 = ∠5 + ∠PQR + ∠6
⇒ ∠2 + ∠3 + ∠1 + ∠4 = ∠2 + ∠PQR + ∠1
[From (iii), ∠5 = ∠2 and ∠6 = ∠1]
⇒ ∠3 + ∠4 = ∠PQR …(vi)

In ΔPQR, we have
∠3 + ∠4 + ∠PQR = 180° (Sum of interior angles of a triangle = 180°)
⇒∠PQR + ∠PQR = 180° [Using (vi)]
⇒ 2∠PQR = 180°
⇒ ∠PQR = $$\frac {180°}{2}$$ = 90°
Hence, ∠PQR = 90°. Hence Proved

Question 9.
ABCD is a square in which P and Q are points on the sides BC and CD respectively such that PQ || BD. If AR bisects the PQ.
prove that :
(i) ∠PAR = ∠QAR
(ii) If AR produce it will passes through C.
Solution :
(i) In ΔBCD, we have
BC = DC, (Square’s sides)
⇒ ∠CBD = ∠CDB
(Angles opposite to equal sides are equal)
Now, PQ || BD (Given) …(i)
∠CPQ = ∠CBD …….(ii)
(Corresponding angles)
∠CQP = ∠CDB, …….(iii)
(Corresponding angles)
From (i), (ii) and (iii), we get
∠CPQ = ∠CQP
⇒ CP = CQ
(Side opposite to equal angles are equal)

Now BC = CD
⇒ BC – CP = CD – CP
⇒ BC – CP = CD – CQ, [∵ CP = CQ]
⇒ BP = DQ ……….(iii)
In ΔABP and ΔADQ, we have
∠ABP = ∠ADQ, (Each = 90°)
and BP = DQ [Using (iii)]
[By SAS congruence rule]
⇒ AP = AQ, (CPCT) …(iv)
Now, in ΔAPR and ΔAQR, we have
AP = AQ, [As proved in (iv)]
AR = AR (Common)
and PR = QR [It is given that R is the midpoint of PQ]
∴ ΔAPR ≅ ΔΑQR,
(By SSS congruence rule)
⇒ ∠PAR = ∠QAR, (CPCT)
Proved

(ii) If AR produce and join to C. It must be a straight line we have PQ || BD.
∠BSR + ∠PRS = 180°,
(Sum of allied angles = 180°)
⇒ ∠PRC + ∠PRS = 180°, [∵ ∠BSR and ∠PRC are corresponding angles ∴ ∠PRC = ∠BSR]
∴ Point A, R and C lies on the line AC.
⇒ AC is a straight line.
Hence, if AR is produced it will passes through C.
Proved

Question 10.
Squares ABCD and AFPQ are drawn on the sides AD and AF of parallelogram ADEF. Prove that BQ = AE.
Solution :
We have
AQ = AF …………(i)
(Square’s sides)
DE = AF …………(ii)
(Opposite sides of a parallelogram)

From (i) And (ii), we get
AQ = DE …………(iii)
⇒ ∠BAQ + 90° + ∠FAD + 90° = 360° (from fig.)
⇒ ∠BAQ + ∠FAD = 360°- 90° – 90°
⇒ ∠BAQ + ∠FAD = 180°
⇒ ∠BAQ = 180° – ∠FAD …(iv)
AF || DE, (Opposite sides of a parallelogram)
(Sum of allied angles = 180°)
From (iv) and (v), we get
Now in ΔBAQ and ΔADE, we have
∠BAQ = ∠ADE, [As proved in (vi)]
and AQ = DE, [As proved in (iii)]
(By SSS congruence rule)
⇒ BQ = AE, (CPCT)
Hence proved.

Question 11.
In the figure, ABC is a triangle in which PQ ⊥ BC and PR ⊥ AB. If BP and CP are the bisectors of ∠B and ∠C respectively. Prove that :
(i) PR = PT,
(ii) AP bisects ∠A.

Solution :
(i) In ΔPRB and ΔPQB, we have
∠PRB = ∠PQB, (Each = 90°)
∠RBP = ∠QBP, [BP is the bisector of ∠B]
and BP = BP, (Common)
∴ ΔPRB ≅ ΔPQB,
(By AAS congruence rule)
⇒ PR = PQ, (CPCT) …(i)
Now draw PT ⊥ AC and join AP.
Again in ΔPQC and ΔPTC, we have
∠PQC = ∠PTC (Each = 90°)
∠PCQ = ∠PCT, (PC is the bisector of ∠C)

and PC = PC (Common)
∴ ΔPQC ≅ ΔPTC,
(By AAS congruence rule)
⇒ PQ = PT, (CPCT)
⇒ PT = PQ …………(ii)
From (i) and (ii), we get
PR = PT ……..(iii)
Hence Proved

(ii) Now in ΔARP and ΔATP, we have
∠ARP = ∠ATP, (Each = 90°)
Hyp. AP = Hyp. AP (Common)
and PR = PT,
[As proved above in (iii)]
∴ ΔARP ≅ ΔATP,
(By RHS congruence rule)
⇒ ∠RAP = ∠TAP (CPCT)
⇒ AP bisects the ∠A.
Hence proved

Question 12.
E is the point on the side AC of ΔABC such that AB = AE. If AD bisects ∠A, prove that :
(i) BD = DE,
(ii) ∠ABE > ∠C.

Solution :
(i) In ΔABE, we have
AB = AE (Given)
⇒ ∠ABE = ∠AEB,
(Angles opposite to equal sides are equal)
⇒ ∠ABP = ∠AEP
In ΔABP and ΔAEP, we have
∠PAB = ∠PAE (Given)
∠ABP = ∠AEP,
[As proved above in (i)]
and AP = AP, (common)
∴ ΔABP ≅ ΔAEP,
(By AAS congruence rule)
⇒ BP = PE (CPCT) …….(ii)
and ∠APB = ∠APE (CPCT)
⇒ ∠EPD = ∠BPD, ………..(iii)
(∵ ∠APB = ∠EPD and ∠APE = ∠BPD)
Now in ΔBPD and ΔEPD, we have
BP = PE, [As proved above in (ii)]
∠BPD = ∠EPD,
[As proved above in (iii)]
and PD = PD, (common)
∴ ΔBPD ≅ ΔΕPD,
(By SAS congruence rule)
⇒ BD = DE (CPCT)
Hence proved

(ii) In ΔBEC, we have
∠AEB = ∠C + ∠EBC, (By theorem 6.8)
∴ ∠AEB > ∠C
⇒ ∠ABE > ∠C, [∵ ∠AEB = ∠ABE]
Hence proved

Question 13.
In the figure, AD, BE and CF are medians of ΔABC. Prove that :
2(AD + BE + CF) > AB + BC + CA.

Solution :
In ΔABD, we have
BD + AD > AB, [∵ Sum of any two sides of a triangle is greater than third side]
⇒ $$\frac {1}{2}$$BC + AD > AB ……..(i)
Again in ΔACD, we have
[∵ Sum of any two sides of a Δ is greater than third side]
⇒ $$\frac {1}{2}$$BC + AD > AC …………(ii)
Adding (i) and (ii), we get
$$\frac {1}{2}$$BC + AD + $$\frac {1}{2}$$BC + AD > AB + AC
⇒ 2AD + BC > AB + AC …………….(iii)
Similarly, we can prove that
2BE + AC > AB + BC ………(iv)
and 2CF + AB > AC + BC ……………(v)
Adding (iii), (iv) and (v), we get
2AD + 2BE + 2CF + AB + BC + AC > AB + AC + AB + BC + AC + BC
⇒ 2AD + 2BE + 2CF > AB + BC + AC
2(AD + BE + CF) > AB + BC + CA.
Hence proved

Question 14.
In the figure, ABC is an equilateral triangle in which BP and CP are bisectors of ∠B and ∠C respectively. If PQ || AB and PR || AC, prove that BQ = QR = RC.

Solution :
We have ΔABC is an equilateral triangle.
∠A = ∠B = ∠C = 60°
∠PBQ = $$\frac {1}{2}$$∠B = $$\frac {60°}{2}$$ = 30°……(i) [∵ BP is the bisector of ∠B]
and ∠PCR = $$\frac {1}{2}$$∠C = $$\frac {60°}{2}$$ = 30°…(ii)
[∵ CP is the bisector of ∠C]
∵ PQ || AB
⇒ ∠PQR = ∠ABQ,
[Corresponding angles]
⇒ ∠PQR = 60° ……….(iii)
[∵ ∠ABQ = ∠ABC = 60°]
and PR || AC
⇒ ∠PRQ = ∠ACR
[Corresponding angles]
⇒ ∠PRQ = 60° ……(iv) [∵ ∠ABR = ∠ACB60°C]
In ΔPQR, we have
∠PQR + ∠PRQ + ∠QPR = 180°, (Sum of interior angles of a triangle = 180°)
⇒ 60° + 60° + ∠QPR = 180°,
[Using (iii) and (iv)]
⇒ ∠QPR = 180° – 60° – 60°
⇒ ∠QPR = 60°
∴ ΔPQR is an equilateral triangle.
⇒ PQ = QR = RP ……..(v)
In ΔPBQ, we have
∠PQR = ∠PBQ + ∠BPQ,
⇒ 60° = 30° + ∠BPQ,
[Using (i) and (iii)]
⇒ ∠BPQ = 60° – 30° = 30°
⇒ ∠PBQ = ∠BPQ, (Each = 30°)
⇒ PQ = BQ …(vi)
Similarly, PR = RC …(vii)
From (v), (vi) and (vii), we get
BQ = QR = RC.
Hence proved

Multiple Choice Questions

Choose the correct option in each of the following:

Question 1.
Which of the following is not a rule ‘for congruence of triangles’ :
(NCERT Exemplar Problems)
(a) SSS
(b) RHS
(c) SSA
(d) SAS
Solution :
(c) SSA

Question 2.
If ΔABC ≅ ΔPQR and ΔABC is not congruent to ΔRPQ, then which of the following is not true :
[NCERT Exemplar Problems]
(a) BC = PQ
(b) AC = PR
(c) QR = BC
(d) AB = PQ
Solution :
(a) BC = PQ

Question 3.
In triangles ABC and DEF, AB = FD and ∠A = ∠D. The two triangles will be congruent by SAS axiom if : [NCERT Exemplar Problems]
(a) BC = EF
(b) AC = DE
(c) AC = EF
(d) BC = DE
Solution :
(b) AC = DE

Question 4.
Two equilateral triangles are congruent, when :
(a) their sides are proportional
(b) their sides are equal
(c) their angles are equal
(d) None of these
Solution :
(b) their sides are equal

Question 5.
ΔABC ≅ ΔPQR, if BC = 4 cm, ∠B = 60° and ∠C = 70°, then which of the following is true :
(a) QR = 4 cm, ∠P = 60°
(b) PQ = 4 cm, ∠Q = 60°
(c) QR = 4 cm, ∠R = 60°
(d) QR = 4 cm, ∠Q = 60°
Solution :
(d) QR = 4 cm, ∠Q = 60°

Question 6.
In figure, ΔABD ≅ ΔACD, AB = AC, BD = CD. Name the criteria by which the triangles are congruent :

(a) SSS
(b) SAS
(c) ASA
(d) RHS
Solution :
(a) SSS

Question 7.
In the figure, ΔABD ≅ ΔACD. Name the criteria by which the triangles are congruent :

(a) SSS
(b) ASA
(c) AAS
(d) RHS
Solution :
(a) SSS

Question 8.
Two right-angled triangles ADB and ADC are congruent if AB = AC. Name the criteria by which the triangles are congruent :
(a) RHS
(b) SAS
(c) ASA
(d) SSA
Solution :
(a) RHS

Question 9.
In the figure, if AB = PQ, BC = QR and AC = PR, then which of the following is true :

(a) ΔABC ≅ ΔRPQ
(b) ΔACB ≅ ΔRQP
(c) ΔABC ≅ ΔPQR
(d) ΔABC ≅ ΔPRQ
Solution :
(a) ΔABC ≅ ΔRPQ

Question 10.
In ΔPQR, ∠R = ∠P and QR = 4 cm and PR = 5 cm. Then the length of PQ is : [NCERT Exemplar Problems]
(a) 4 cm
(b) 5 cm
(c) 2 cm
(d) 2.5 cm
Solution :
(a) 4 cm

Question 11.
In ΔABC, BC = AB and ∠B = 80°. Then ∠A is equal to [NCERT Exemplar Problems]
(a) 80°
(b) 40°
(c) 50°
(d) 100
Solution :
(c) 50°

Question 12.
In triangles ABC and PQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are:
[NCERT Exemplar Problems]
(a) isosceles but not congruent
(b) isosceles and congruent
(c) congruent but not isosceles
(d) neither congruent nor isosceles
Solution :
(a) isosceles but not congruent

Question 13.
In the figure, PQ = PR and QS = RS, then ratio of ∠PQS : ∠PRS is :

(a) 1 : 2
(b) 2 : 1
(c) 2 : 3
(d) 1 : 1
Solution :
(d) 1 : 1

Question 14.
In a ΔABC, we have:
(a) AB + BC > AC
(b) AB + AC = BC
(c) AB + BC < AC
(d) None of these
Solution :
(a) AB + BC > AC

Question 15.
In ΔABC, we have :
(a) AB + BC < AC
(b) AC – AB < BC (c) BC – AC > AB
(d) BC – AB > AC
Solution :
(b) AC – AB < BC