# Frank Solutions for Chapter 12 Isosceles Triangles Class 9 Mathematics ICSE

**Exercise 12.1**

**1. Find the angles of an isosceles triangle whose equal angles and the non-equal angles are in the ratio 3:4.**

**Answer:**

The equal angles and the non-equal angles are in the ratio 3:4

Let the equal angles be 3x each,

So, non- equal angle is 4x

We know that,

Sum of angles of a triangle = 180°

Hence,

3x + 3x + 4x = 180°

⇒ 10x = 180°

We get,

x = 18°

Therefore,

3x = 3 × 18° = 54° and

4x = 4 × 18° = 72°

Hence,

The angles of a triangle are 54°, 54° and 72°

**2. Find the angles of an isosceles triangle which are in the ratio 2:2:5**

**Answer**

Let equal angles be 2x each

So, non-equal angle is 5x

We know that,

Sum of angles of a triangle = 180°

2x + 2x + 5x = 180°

9x = 180°

x = 20°

Therefore,

2x = 2 × 20° = 40°

5x = 5 × 20° = 100°

Hence, the angles of a triangle are 40°, 40° and 100°

**3. Each equal angle of an isosceles triangle is less than the third angle by 15°. Find the angles.**

**Answer**

Therefore, non-equal angle is x + 15°

We know that,

Sum of angles of a triangle = 180°

x + x + (x + 15°) = 180°

⇒ 3x + 15° = 180°

⇒ 3x = 180° – 15°

⇒ 3x = 165°

We get,

x = 55°

So,

(x + 15°) = 55° + 15° = 70°

Hence, the angles of a triangle are 55°, 55° and 70°

**4. Find the interior angles of the following triangles**

**(a)**

**(b)**

**(c)**

**(d)**

**Answer**

**(a)**

∠A = 110°

AB = AC

∠C = ∠B **(angles opposite to two equal sides are equal)**

Now,

By angle sum property,

∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠B + ∠B = 180°

⇒ 110° + 2∠B = 180°

⇒ 2∠B = 180° – 110°

⇒ 2∠B = 70°

We get,

∠B = 35°

∠C = 35°

Hence,

The interior angles are ∠B = 35° and ∠C = 35°

**(b)**

In △ABC,

AB = AC

∠ACB = ∠ABC **...(1) [∵ angles opposite to two equal sides are equal]**

Now,

∠ACB + ∠ACD = 180° **[linear pair]**

∠ACB = 180° – ∠ACD

∠ACB = 180° – 105°

∠ACB = 75°

So,

∠ABC = 75° [from equation (1)]

Now, in △ABC,

By angle sum property,

∠ABC + ∠ACB + ∠BAC = 180°

75° + 75° + ∠BAC = 180°

150° + ∠BAC = 180°

∠BAC = 180° – 150°

We get,

∠BAC = 30°

Hence,

In △ABC,

∠A = 30°

∠B = 75°

∠C = 75°

**(c)**

In △ABD,

Given that,

AD = BD

∠ABD = ∠BAD **…(angles opposite to two equal sides are equal)**

Now,

∠ABD = 37° **…(given)**

Hence,

∠BAD = 37°

By exterior angle property,

∠ADC = ∠ABD + ∠BAD

∠ADC = 37° + 37°

We get,

∠ADC = 74°

In △ADC,

AC = DC **…(given)**

∠ADC = ∠DAC **…(angles opposite to two equal sides are equal)**

∠DAC = 74°

Now,

∠BAC = ∠BAD + ∠DAC

∠BAC = 37° + 74°

We get,

∠BAC = 111°

In △ABC,

∠BAC + ∠ABC + ∠ACB = 180°

111° + 37° + ∠ACB = 180°

∠ACB = 180° – 111° – 37°

We get,

∠ACB = 32°

Therefore,

The interior angles of △ABC are 37°, 111° and 32°

**(d)**

AD = CD **…(given)**

∠ACD = ∠CAD **…(angles opposite to two equal sides are equal)**

Now,

∠ACD = 50° **…(given)**

∠CAD = 50°

By exterior angle property,

∠ADB = ∠ACD + ∠CAD

∠ADB = 50^{0} + 50^{0}

∠ADB = 100^{0}

In △ADB,

AD = BD **…(given)**

∠DBA = ∠DAB **…(angles opposite to two equal sides are equal)**

Also,

∠ADB + ∠DBA + ∠DAB = 180°

⇒ 100° + 2∠DBA = 180°

⇒ 2∠DBA = 180° – 100°

⇒ 2∠DBA = 80°

We get,

∠DBA = 40°

∠DAB = 40°

∠BAC = ∠DAB + ∠CAD

⇒ ∠BAC = 40° + 50°

⇒ ∠BAC = 90°

Therefore, the interior angles of △ABC are 50°, 90° and 40°

**5. Side BA of an isosceles triangle ABC is produced so that AB = AD. If AB and AC are the equal sides of the isosceles triangle, prove that ****∠****BCD is a right angle.**

**Answer:**

Let ∠ABC = x

Hence,

∠BCA = x (since AB = AC)

In △ABC,

∠ABC + ∠BCA + ∠BAC = 180° **…(1)**

But

∠BAC + ∠DAC = 180° **…(2)**

From equations (1) and (2)

∠ABC + ∠BCA + ∠BAC = ∠BAC + ∠DAC

⇒ ∠DAC = ∠ABC + ∠BCA

⇒ ∠DAC = x + x

We get,

∠DAC = 2x

Let ∠ADC = y,

Hence,

∠DCA = y **(since AD = AC)**

Now,

In △ADC,

∠ADC + ∠DCA + ∠DAC = 180° **…(3)**

But ∠BAC + ∠DAC = 180° **…(4)**

From equations (3) and (4), we get,

∠ADC + ∠DCA + ∠DAC = ∠BAC + ∠DAC

⇒ ∠BAC = ∠ADC + ∠DCA

⇒ ∠BAC = y + y

⇒ ∠BAC = 2y

Now, substituting the value of ∠BAC and ∠DAC in equation (2)

2x + 2y = 180°

⇒ x + y = 90°

⇒ ∠BCA + ∠DCA = 90°

Therefore,

∠BCD is a right angle

**6. The bisectors of the equal angles of an isosceles triangle PQR meet at O. If PQ = PR, prove that PO bisects ****∠****P.**

**Answer**

Join PO and produce to meet QR at point S

In △PQS and △PRS

PS = PS **(common)**

PQ = PR **(given)**

So,

∠Q = ∠R **(angles opposite to two equal sides are equal)**

Hence,

△PQS ≅ △PRS

Thus,

∠QPS = ∠RPS

Therefore,

PO bisects ∠P

**7. Prove that the medians corresponding to equal sides of an isosceles triangle are equal.**

**Answer**

Let D and E be the mid points of AB and AC respectively

Now,

Join BE and CD

Then BE and CD become the medians of this isosceles triangle

In △ABE and △ACD

AB = AC **(given)**

AD = AE **(D and E are mid points of AB and AC)**

∠A = ∠A **(common angle)**

Hence,

△ABE ≅ △ACD **(SAS criteria)**

Therefore,

The medians BE and CD are equal i.e. BE = CD

**8. DPQ is an isosceles triangle with DP = DQ. A straight line CD bisects the exterior ****∠****QDR. Prove that DC is parallel to PQ**

**Answer**

In △QDP,

DP = DQ

Hence,

∠Q = ∠P **(angles opposite to two equal sides are equal)**

∠QDR = ∠Q + ∠P

⇒ 2∠QDC = ∠Q + ∠P **(DC bisects angle QDR)**

⇒ 2∠QDC = ∠Q + ∠Q

We get,

2∠QDC = 2∠Q

Hence,

∠QDC = ∠Q

But these angles are alternate angles

Therefore,

DC || PQ

Hence, proved

**9. In a quadrilateral PQRS, PQ = PS and RQ = RS. If ****∠****P = 50**° **and ****∠****R = 110**°**, find ****∠****PSR.**

**Answer**

In △PQS,

PQ = PS

Therefore,

∠PQS = ∠PSQ **(angles opposite to two equal sides are equal)**

∠P + ∠PQS + ∠PSQ = 180°

⇒ 50° + 2∠PQS = 180°

⇒ 2∠PQS = 180° – 50°

We get,

2∠PQS = 130°

⇒ ∠PQS = 65°

So,

∠PQS = ∠PSQ = 65° **…(1)**

In △SRQ,

SR = RQ

Hence,

∠RQS = ∠RSQ (angles opposite to two equal sides are equal)

∠R + ∠RQS + ∠RSQ = 180°

⇒ 110° + 2∠RQS = 180°

⇒ 2∠RQS = 180° – 110°

We get,

2∠RQS = 70°

∠RQS = 35°

So,

∠RQS = ∠RSQ = 35° **….(2)**

Adding equations (1) and (2), we get,

∠PSQ + ∠RSQ = 65° + 35°

∠PSR = 100°

**10. △****ABC is an isosceles triangle with AB = AC. Another triangle BDC is drawn with base BC = BD in such a way that BC bisects ****∠****B. If the measure of ****∠****BDC is 70**°**, find the measures of ****∠****DBC and ****∠****BAC.**

**Answer**

In △BDC,

∠BDC = 70°

BD = BC

Hence,

∠BDC = ∠BCD **(angles opposite to two equal sides are equal)**

∠BCD = 70°

Now,

∠BCD + ∠BDC + ∠DBC = 180°

⇒ 70° + 70° + ∠DBC = 180°

⇒ ∠DBC = 180° – 140°

We get,

∠DBC = 40°

∠DBC = ∠ABC **(BC is the angle bisector)**

Hence,

∠ABC = 40°

In △ABC,

Since AB = AC, ∠ABC = ∠ACB

Hence,

∠ACB = 40°

⇒ ∠ACB + ∠ABC + ∠BAC = 180°

⇒ 40° + 40° + ∠BAC = 180°

⇒ ∠BAC = 180° – 80°

⇒ ∠BAC = 100°

Therefore, the measure of ∠BAC = 100° and ∠DBC = 40°

**11. △****PQR is isosceles with PQ = PR. T is the mid-point of QR, and TM and TN are perpendiculars on PR and PQ respectively. Prove that,**

**(i) TM = TN**

**(ii) PM = PN and**

**(iii) PT is the bisector of ****∠****P**

**Answer**

(i) In △PQR,

PQ = PR

Hence,

∠R = ∠Q **…(1)**

Now,

In △QNT and △RMT

∠QNT = ∠RMT = 90°

∠Q = ∠R **[from equation (1)]**

QT = TR** (given)**

Hence,

△QNT ≅ △RMT **(AAS criteria)**

Therefore,

TM = TN

**(ii)** Since, △QNT ≅ △RMT

NQ = MR **…(2)**

But,

PQ = PR **…(3) [given]**

Now, subtracting (2) from (3), we get,

PQ – NQ = PR – MR

PN = PM

**(iii)** In △PNT and △PMT

TN = TM **(proved)**

PT = PT **(common)**

∠PNT = ∠PMT = 90°

Hence,

△PNT ≅ △PMT

So,

∠NPT = ∠MPT

Therefore,

PT is the bisector of ∠P

**12. △****PQR is isosceles with PQ = QR. QR is extended to S so that ****△****PRS becomes isosceles with PR = PS. Show that ****∠****PSR: ****∠****QPS = 1:3**

**Answer**

PQ = QR **(given)**

∠PRQ = ∠QPR ** …(1)**

In △PRS,

PR = RS (given)

∠PSR = ∠RPS **…(2)**

Now,

Adding equations (1) and (2), we get,

∠QPR + ∠RPS = ∠PRQ + ∠PSR

∠QPS = ∠PRQ + ∠PSR **…(3)**

Now,

In △PRS,

∠PRQ = ∠RPS + ∠PSR

∠PRQ = ∠PSR + ∠PSR **[from equation(2)]**

∠PRQ = 2∠PSR **…(4)**

Now,

∠QPS = 2∠PSR + ∠PSR **[from equation (3) and (4)]**

∠QPS = 3∠PSR

∠PSR/∠QPS = 1/3

Therefore,

∠PSR: ∠QPS = 1: 3

Hence, proved

**13. In ****△****KLM, KT bisects ****∠****LKM and KT = TM. If ****∠****LTK is 80 ^{0}, find the value of **

**∠**

**LMK and**

**∠**

**KLM.**

**Answer**

In △KTM,

KT = TM (given)

Hence,

∠TKM = ∠TMK **…(1)**

Now,

∠KTL = ∠TKM + ∠TMK

⇒ 80° = ∠TKM + ∠TKM **…[from (1)]**

⇒ 80° = 2∠TKM

We get,

∠TKM = 40° = ∠TMK = ∠LMK **…(2)**

But,

∠TKM = ∠TKL **(KT is the angle bisector)**

Hence,

∠TKL = 40°

In △KTL,

∠TKL + ∠KTL + ∠KLT = 180°

⇒ 40° + 80° + ∠KLT = 180°

⇒ ∠KLT = 180° – 40° – 80°

We get,

∠KLT = 60^{0} = ∠KLM

Therefore,

∠KLM = 60° and ∠LMK = 40°

**14. Equal sides QP and RP of an isosceles ****△****PQR are produced beyond P to S and T such that ****△****PST is an isosceles triangle with PS = PT. Prove that TQ = SR.**

**Answer**

In △PTQ and △PSR

PQ = PR **(given)**

PT = PS **(given)**

∠TPQ = ∠SPR **(vertically opposite angles)**

Hence,

△PTQ ≅ △PSR

Therefore,

TQ = SR

Hence, proved

**15. Prove that the bisector of the vertex angle of an isosceles triangle bisects the base perpendicularly.**

**Answer**

AB = AC **(given)**

AD = AD **(common)**

∠BAD = ∠CAD **(AD bisects ∠BAC)**

Hence,

△ADB ≅ △ADC

Therefore,

BD = DC and ∠BDA = ∠CDA

But,

∠BDA + ∠CDA = 180°

∠BDA = ∠CDA = 90°

Therefore,

AD bisects BC perpendicularly

Hence, proved

**16. In the figure ****△ABC is isosceles with AB = AC. Prove that: **

**(i) ∠A : ∠B = 1 : 3 **

**(ii) ∠ADE = ∠BCD **

**Answer**

In △DEC,

∠DEC = ∠ADE + ∠A = 2a **(ext. Angle to △ADE)**

DE = DC

⇒ ∠DEC = ∠DEC = 2a **…(ii)**

In △BDC, let ∠B = b

DC = BC

⇒ ∠BDC = ∠B = b **…(iii)**

In △ABC,

∠ADB = ∠ADE + ∠EDC + ∠BDC

180° = a + ∠EDC + b **(from (i) and (iii))**

∠EDC = 180° - a – b **…(iv) **

Now again in △DEC

180° = ∠EDC + ∠DCE + ∠DEC **(from (ii))**

⇒ 180° = ∠EDC + 2a + 2a

⇒ ∠EDC = 180° - 4a **…(v) **

Equating (iv) and (v)

180° - a – b = 180° - 4a

⇒ 3a = b **…(vi)**

⇒ a/b = 1/3 = ∠A/∠B

Hence, ∠A : ∠B = 1 : 3

**(ii)** In △ADE,

∠ADE = a **(from (i))**

In △BCD,

∠BCE = b (AB = AC ⇒ ∠B = ∠C) **…(vii)**

⇒ ∠BCD = ∠BCE - ∠DCE

⇒ ∠BCD = b – 2a **(from (vii) and (ii))**

But b = 3a **(from (vi))**

Therefore, ∠BCD = 3a – 2a = a

Hence, ∠ADE = ∠BCD

**17. In ****△ABC, D is mid-point of BC, AD is equal to AC. AC is produced to E, such that CE = AC. Prove that : **

**(i) ∠ADB = ∠DCE **

**(ii) AB = CE **

**Answer**

**(i)** In △ADC,

AD = AC (given)

Therefore, ∠ADC = ∠ACD **…(i)**

But ∠ADB + ∠ADC = 180° **…(ii) **

And ∠ACD + ∠DCE = 180° **…(iii) **

From (i), (ii) and (iii)

∠ADB = ∠DCE

**(ii)** In △ABD and △DCE

BD = CD **(D is mid-point of BC)**

∠ADB = ∠DCE **(proved)**

AD = CE **(since AC = AC and AC = CE)**

Therefore, △ABD ≅ △DCE

Hence, AB = CE

**18. In △XYZ, AY and AZ are the bisectors of ∠Y and ∠Z respectively. The perpendicular bisectors of AY and AZ cut YZ at B and C respectively. Prove that line segment YZ is equal to the perimeter of △ABC. **

**Answer**Let M and N be the points where AY and AZ are bisected.

In △ABM and △BMY

MY = MA **(BM bisects AY)**

BM = BM **(common)**

∠BMY = ∠BMA

Therefore, △ABM ≅ △BMY

Hence, YB = AB **…(i)**

In △ACN and △CNZ

NZ = NA **(CN bisects AZ)**

CN = CN **(common)**

Therefore, △ACN ≅ △CNZ

Hence, CZ = AC **…(ii)**

YZ = YB + BC + CZ

Substituting from (i) and (ii)

YZ = YB + BC + AC

Hence, YZ is equal to the perimeter of △ABC

**19. △PQR is an isosceles triangle with PQ = PR. QR is extended to S and ST is drawn perpendicular to QP produced, and SN is perpendicular to PR produced. Prove that QS bisects ∠TSN. **

**Answer**

In △PQR, let ∠PQR = x

PQ = PR

⇒ ∠PQR = ∠PRQ = x **…(i)**

In △RNS,

∠NRS = ∠PRQ = x **(vertically opposite angles)**

∠RNS = 90**° **(given)

⇒ ∠NSR + ∠RNS + ∠NRS = 180°

⇒ ∠NSR = 90° - x **…(ii) **

Now in Quadrilateral PTRS

∠PTS = 90° **(given) **

⇒ ∠TPR = ∠PQR + ∠PRQ = 2x **(exterior angle to triangle PQR)**

⇒ ∠PRS = 180° - ∠PRQ = 180° - x **(QRS is a st. Line)**

⇒ ∠PTS + ∠TPR + ∠PRS + ∠TSR = 360° (angles of a quad. = 360°)

⇒ 90° + 2x + 180° - x + ∠TSR = 360°

⇒ ∠TSR = 90° - x **…(iii) **

From (ii) and (iii)

∠TSR = ∠NSR

Therefore, QS bisects ∠TSN.

**20. In the given figure, D and E are points on AB and AC respectively. AE and CD intersect at P such that AP = CP. If ∠BAE = **∠**BCD, prove that DBDE is isosceles****. **

**Answer**Join DE and AC,

In △APD and △EPC,

∠DAP = ∠ECP **...(∵ ∠BAE = ∠BCD)**

AP = CP **…(given)**

∠APD = ∠EPC **…(vertically opposite angles)**

∴ △APD ≅ △EPC **…(By ASA Congruence criterion)**

⇒ AD = EC **…(c.p.c.t)**

In △APC,

AP = CP **…(given) **

⇒ ∠PAC = ∠PCA **…(angles opposite to two equal sides are equal)**

Now, ∠BAE = ∠BCD and ∠PAC = ∠PCA

⇒ ∠BAC = ∠BCA

⇒ BC = BA **...(sides opposite to two equal angles are equal)**

⇒ BE + EC = BD + DA

⇒ BE = BD **(∵ EC = DA)**

⇒ ∠BDE = ∠BED **…(angles opposite to two equal sides are equal)**

⇒ △BDE is an isosceles triangle.

**21. In DPQR as shown, ∠PQS = ∠RQS and QS ****⊥ PR. Find the value of x and y, if PQ = 3x + 1; QR = 5y – 2; PS = x + 1 and SR = y + 2. **

**Answer**In △PQS and △SQR,

QS = QS **…[common]**

∠QSP = ∠QSR **…[each = 90°]**

∠PQS = ∠RQS **…[given]**

∴ △PQS ≅ △SQR **…[By ASA criterion]**

⇒ PS = RS

⇒ x + 1 = y + 2

⇒ x = y + 1 **…(i)**

And PQ = SQ

⇒ 3x + 1 = 5y – 2

⇒ 3(y + 1) + 1 = 5y – 2 **…[From (i)]**

⇒ 3y + 3 + 1 = 5y – 2

⇒ 2y = 6

⇒ y = 3

Putting y = 3 in (i),

x = y + 1 = 3 + 1 = 4

**22. If ****∠****PQS = 60****°,**

**a. find ****∠****QPR.**

**b. show that PQ = PS = QS = SR**

**Answer**

**(a) **In △PQS,

PQ = QS **…(given)**

⇒ ∠QSP = ∠QPS **…(angles opposite to two equal sides are equal)**

Now, ∠PQS + ∠QSP + ∠QPS = 180**°**

⇒ 60**° ****+** 2∠QSP = 180**°**

⇒ 2∠QSP = 120 **°**

⇒ ∠QSP = 60**°**

⇒ QPS = 60**°**

In △PRS,

PS = SR **…(given)**

⇒ ∠PRS = ∠RPS **…(angles opposite to two equal sides are equal)**

By exterior angle property,

∠QSP = ∠RPS + ∠PRS

⇒ 60**° = **2∠RPS

⇒ ∠RPS = 30**°**

Now ∠QPR = ∠QPS + ∠RPS = 60**° + **30° = 90°

**(b) **In △PQS,

∠PQS = 60**°, **∠QPS = 60**°** and ∠QSP = 60**° **

⇒ △PQS is an equilateral triangle.

⇒ PQ = QS = PS

And, PS = SR

⇒ PQ = PS = QS = SR

**23. In the given figure, if DPQR is an isosceles triangle, prove that : ****∠****QSR = exterior ****∠****PRT. **

**Answer**Let ∠PQS = SQR = x and ∠PRS = ∠SRQ = y

In △PQR,

∠QPR + ∠PQR + ∠PRQ = 180**°**

⇒ ∠QPR + 2x + 2y = 180**°**

⇒ ∠QPR + 2x + 2y = 180**°**

⇒ ∠QPR = 180**° ****- **2x – 2y **…(i)**

Since PQ = PR,

∠PRQ = ∠PQR **…(angles opposite to two equal sides are equal)**

⇒ 2x = 2y

⇒ x = y

Now, ∠PRT = ∠PQR + ∠QPR **…(by exterior angle property)**

⇒ ∠PRT = 2x + 180**° -** 2x – 2y **…[From (i)]**

⇒ ∠PRT = 180° - 2y **…(ii) **

In △SQR,

∠QSR + ∠SQR + ∠SRQ = 180

⇒ ∠QSR + x + y = 180**°**

⇒ ∠QSR = 180**° -** x – y

⇒ ∠QSR = 180° - y – y** ….[∵ x = y (proved)]**

⇒ ∠QSR = 180° - 2y** …(iii)**

From (ii) and (iii),

∠QSR = ∠PRT

**24. In the given figure, if DABC is an isosceles triangle and ****∠****PAC = 110****°, find the base angle and vertex angle of the DABC. **

**Answer**Given : ∠PAC = 110

**°**

To find:

Base angles :** **∠ABC and ∠ACB

Vertex angle: ∠BAC

In quadrilateral APQC,

∠APQ + ∠PQC + ∠ACQ + ∠PAC = 360**°**

⇒ 90**°** + 90**°** + ∠QCA + 110**°** = 360**°**

⇒ ∠ACQ = 360**° -** 290°

⇒ ∠ACQ = 70°

⇒ ∠ACB = 70° **…(i)**

In △ABC,

AB = AC **…(given)**

⇒ ∠ACB = ∠ABC **...(angles opposite to two equal sides are equal)**

⇒ ∠ABC = 70° **…[From (i)]**

In △ABC,

∠ABC + ∠ACB + ∠BAC = 180° **…(angle sum property)**

⇒ 70**° +** 70° **+ **∠BAC = 180**° **

⇒ ∠BAC = 180**° **- 140**°**

⇒ ∠BAC = 40**°**