# Frank Solutions for Chapter 11 Triangles and their Congruency Mathematics ICSE

**Exercise 11.1**

**1. In the given figure, ****∠****Q: ****∠****R = 1: 2. Find:**

**(a) ****∠****Q**

**(b) ****∠****R**

**Answer**Given,

∠Q: ∠R = 1: 2

Let us consider ∠Q = x°

∠R = 2x°

Now,

∠RPX = ∠Q + ∠R **[by exterior angle property]**

105° = x° + 2x°

⇒ 105° = 3x°

We get,

x° = 35°

Therefore,

∠Q = x° = 35° and

∠R = 2x° = 70°

**2. The exterior angles, obtained on producing the side of a triangle both ways, are 100**° **and 120**°**. Find all the angles of the triangle.**

**Answer**

100° + ∠ABC = 180°

⇒ ∠ABC = 180° – 100°

⇒ ∠ABC = 80°

∠ACQ + ∠ACB = 180° **(Linear pair)**

⇒ 120° + ∠ACB = 180°

⇒ ∠ACB = 180° – 120°

⇒ ∠ACB = 60°

Now,

In △ABC,

∠A + ∠B + ∠C = 180° (Angle sum property of a triangle)

∠A + 80° + 60° = 180°

∠A = 180° – 80° – 60°

We get,

∠A = 40°

Therefore, the angles of a triangle are 40°, 60° and 80°

**3. Use the given figure to find the value of x in terms of y. Calculate x, if y = 15**°**.**

**Answer**(2x – y)° = (x + 5°) + (2y + 25)°

**(Exterior angle property)**

⇒ 2x° – y° = x° + 5° + 2y° + 25°

⇒ 2x° – x° = 2y° + y° + 30°

⇒ x° = 3y° + 30°

When y = 15°,

We have,

x° = 3 × 15° + 30°

⇒ x° = 45° + 30°

We get,

x° = 75°

**4. In a triangle PQR, ****∠****P + ****∠****Q = 130****° ****and ****∠****P + ****∠****R = 120****°****. Calculate each angle of the triangle.**

**Answer**

In △PQR,

∠P + ∠Q = 130°

WKT

∠P + ∠Q = ∠PRY **(Exterior angle property)**

∠PRY = 130°

∠PRY + ∠R = 180° **(Linear pair)**

⇒ 130° + ∠R = 180°

⇒ ∠R = 180° – 130°

We get,

∠R = 50°

Also,

Given

∠P + ∠R = 120°

Now,

∠P + ∠R = ∠PQX **(Exterior angle property)**

∠PQX = 120°

∠PQX + ∠Q = 180° **(Linear pair)**

⇒ 120° + ∠Q = 180°

∠Q = 180° – 120°

We get,

∠Q = 60°

In △PQR,

∠P + ∠Q + ∠R = 180° **(Angle sum property of a triangle)**

⇒ ∠P + 60° + 50° = 180°

⇒ ∠P = 180° – 110°

We get,

∠P = 70°

Therefore, the angles of △PQR are,

∠P = 70°

∠Q = 60° and

∠R = 50°

**5. The angles of a triangle are (x + 10)****°****, (x + 30)****°****and (x – 10)****°****. Find the value of ‘x’. Also, find the measure of each angle of the triangle.**

**Answer**

For any triangle,

Sum of measures of all three angles = 180°

Hence,

We have,

(x + 10)° + (x + 30)° + (x – 10)° = 180°

⇒ x° + 10° + x° + 30° + x° – 10° = 180°

⇒ 3x° + 30° = 180°

⇒ 3x° = 180° – 30°

We get,

3x° = 150°

⇒ x° = 50°

Now,

(x + 10)° = (50 + 10)°

⇒ (x + 10)° = 60°

⇒ (x + 30)° = (50 + 30)°

⇒ (x + 30)° = 80°

⇒ (x – 10)° = (50 – 10)°

⇒ (x – 10)° = 40°

Therefore, the angles of a triangle are 60°, 80° and 40°

**6. Use the given figure to find the value of y in terms of p, q and r**

**Answer**

Here, SR is produced to meet PQ at point E

In △PSE,

∠P + ∠S + ∠PES = 180° **(Angle sum property of a triangle)**

⇒ p° + y° + ∠PES = 180°

⇒ ∠PES = 180° – p° – y° **…(1)**

In △RQE,

∠R + ∠Q + ∠REQ = 180° **(Angle sum property of a triangle)**

⇒ (180° – q°) + r° + ∠REQ = 180°

⇒ ∠REQ = 180° – (180° – q°) – r°

⇒ ∠REQ = q° – r° **…(2)**

Now,

∠PES + ∠REQ = 180° **(Linear pair)**

(180°– p° – y°) + (q° – r°) = 180° **[from (1) and (2)]**

⇒ -p° – y° + q° – r° = 0

⇒ –y° = -q° + p° + r°

We get,

y° = q° – p° – r°

**7. In the figure given below, if RS is parallel to PQ, then find the value of ****∠****y.**

**Answer**

∠P + ∠Q + ∠R = 180° **(angle sum property)**

⇒ 4x° + 5x° + 9x° = 180°

⇒ 18x° = 180°

⇒ x = 10

∠P = 4x° = 4 × 10°

∠P = 40°

∠Q = 5x° = 5 × 10°

⇒ ∠Q = 50°

∠QPR = ∠PRS **(Alternate angles)**

And,

∠QPR = 40°

∠PRS = 40°

By exterior angle property,

∠PQR + ∠QPR = ∠PRS + y°

⇒ 40° + 50° = 40° + y°

We get,

y = 50°

**8. Use the given figure to show that: ****∠****p + ****∠****q + ****∠****r = 360**°

**Answer**

By exterior angle property,

∠p = ∠PQR + ∠PRQ

∠q = ∠QPR + ∠PRQ

∠r = ∠PQR + ∠QPR

Now,

∠p + ∠q + ∠r = ∠PQR + ∠PRQ + ∠QPR + ∠PRQ + ∠PQR + ∠QPR

On further calculation, we get,

∠p + ∠q + ∠r = 2 ∠PQR + 2∠PRQ + 2 ∠QPR

⇒ ∠p + ∠q + ∠r = 2 (∠PQR + ∠PRQ + ∠QPR)

⇒ ∠p + ∠q + ∠r = 2 × 180° (Angle Sum property: ∠PQR + ∠PRQ + ∠QPR = 180°)

We get,

∠p + ∠q + ∠r = 360°

Hence,

∠p + ∠q + ∠r = 360°

**9. In ****△****ABC and ****△****PQR and, AB = PQ, BC = QR and CB and RQ are extended to X and Y respectively and ****∠****ABX = ****∠****PQY. Prove that ****△****ABC ****≅** **△****PQR.**

**Answer**

In △ABC and △PQR

AB = PQ

BC = QR

∠ABX + ∠ABC = ∠PQY + ∠PQR = 180°

So,

∠ABX = ∠PQY

∠ABC = ∠PQR

Therefore,

△ABC ≅ △PQR **(SAS criteria)**

Hence, proved

**10. In the figure, ****∠****CPD = ****∠****BPD and AD is the bisector of ****∠****BAC. Prove that ****△****CAP ****≅**** BAP and CP = BP.**

**Answer**

In △BAP and △CAP

∠BAP = ∠CAP (AD is the bisector of ∠BAC)

AP = AP

∠BPD + ∠BPA = ∠CPD + ∠CPA = 180°

We get,

∠BPD = ∠CPD

∠BPA = ∠CPA

Therefore,

△CAP ≅ △BAP **(ASA criteria)**

So,

CP = BP

Hence, proved

**11. In the figure, BC = CE and ****∠****1 = ****∠****2. Prove that ****△****GCB ****≅** **△****DCE.**

**Answer**

In △GCB and △DCE

∠1 + ∠GBC = ∠2 + ∠DEC = 180°

⇒ ∠1 = ∠2

∠GBC = ∠DEC

So,

BC = CE

∠GCB = ∠DCE **(vertically opposite angles)**

Therefore,

△GCB ≅ △DCE **(ASA criteria)**

Hence, proved.

**12. In the figure, AB = EF, BC = DE, AB and FE are perpendicular on BE. Prove that ****△****ABD ****≅** **△****FEC**

**Answer**

Given that,

In △ABD and △FEC

AB = FE and

BD = CE **(∵ BC = DE; CD is common)**

Therefore,

∠B = ∠E

△ABD ≅ △FEC **(SAS criteria)**

Hence, proved

**13. In the figure, BM and DN are both perpendiculars on AC and BM = DN. Prove that AC bisects BD.**

**Answer**

In △BMR and △DNR

BM = DN

∠BMR = ∠DNR = 90°

∠BRM = ∠DRN **(vertically opposite angles)**

Hence,

∠MBR = ∠NDR **(Sum of angles of a triangle = 180**°**)**

△BMR ≅ △DNR **(ASA criteria)**

Therefore,

BR = DR

So,

AC bisects BD

Hence, proved

**14. AD and BE are altitudes of an isosceles triangle ABC with AC = BC. Prove that AE = BD.**

**Answer:**

In △CAD and △CBE

CA = CB **(Isosceles triangle)**

∠CDA = ∠CEB = 90°

∠ACD = ∠BCE **(common)**

Therefore,

△CAD ≅ △CBA **(AAS criteria)**

Hence,

CE = CD

But,

CA = CB

AE + CE = BD + CD

AE = BD

Hence, proved

**15. In ****△****ABC, X and Y are two points on AB and AC such that AX = AY. If AB = AC, prove that CX = BY.**

**Answer**

In △ABC

AB = AC

AX = AY

BX = CY

In △BXC and △CYB

BX = CY

BC = BC

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

Therefore,

△BXC ≅ △CYB **(SAS criteria)**

So,

CX = BY

Hence, proved

**Exercise 11.2**

**1. Which of the following pairs of triangles are congruent? Give reasons**

**(i) ∆ABC; (BC = 5 cm, AC = 6 cm, ****∠****C = 80°); **

** ∆XYZ; (XZ = 6 cm, XY = 5 cm, ****∠****X = 70°).**

**(ii) ****∆ABC; (AB = 8 cm, BC = 6 cm, ****∠****B = 100°)**

**∆PQR; (PQ = 8 cm, RP = 5 cm, ****∠****Q = 100°)**

**(iii) ****∆ABC; (AB = 5 cm, BC = 7 cm, CA = 9 cm); **

**∆KLM; (KL = 7 cm, LM = 5 cm, KM = 9 cm).**

**(iv) ∆ABC; (****∠****B = 70°, BC = 6 cm, ****∠****C = 50°);**

**∆XYZ; (****∠****Z = 60°, XY = 6 cm, ****∠****X = 70°);**

**(v) ****∆ABC; (****∠****B = 60°, BC = 6 cm, AB = 8 cm);**

**∆PQR; (****∠****Q = 60°, PQ = 6 cm, PR = 10 cm).**

**Answer**

**(i) **

AC = XZ

BC = XY

The included angle ∠C = 80° is not equal to ∠X i.e. 70°.

Now, for ∆ABC to be congruent to ∆XYZ, AB should be equal to XY and YZ should be equal to BC. Then, ∠A = ∠C and ∠X = ∠Z. So, the measure of ∠B will not be equal to ∠Y.

**(ii)**

AB = PQ

∠B = ∠Q

BC can be equal to QR or AC can be equal to RP

Therefore,

∆ABC can be congruent to ∆PQR.

**(iii)**

In ∆ABC and ∆KLM

AB = LM

BC = KL

AC = KM

Therefore,

∆ABC ≅ ∆KLM **(SSS criteria)**

**(iv)**

In ∆ABC and ∆XYZ

∠B = ∠X

BC = XY

Y = 180**° - **(70° + 60°) = 50°

∠C = ∠Y

Therefore,

∆ABC ≅ XYZ **(ASA criteria)**

**(v)**

In ∆ABC and ∆PQR

∠B = ∠Q

BC = PQ

By Pythagoras theorem,

**2. In ∆ABC and ∆PQR and, AB = PQ, BC = QR and CB and RQ are extended to X and Y respectively and ****∠****ABX = ****∠****PQY. Prove that ∆ABC ****≅** **∆PQR.**

**Answer**

In ∆ABC and ∆PQR and

AB = PQ

BC = QR

∠ABX + ∠ABC = ∠PQY + ∠PQR = 180**°**

∠ABX = ∠PQY

⇒ ∠ABC = ∠PQR

Therefore,

∆ABC ≅ ∆PQR **(SAS criteria)**

**3. In the figure, ****∠****CPD = ****∠****BPD and AD is the bisector of ****∠****BAC. Prove that ∆CAP ****≅** **∆BAP and CP = BP. **

**Answer**

In ∆BAP and ∆CAP

∠BAP = ∠CAP **(AD is the bisector of ∠BAC)**

AP = AP

∠BPD + ∠BPA = ∠CPD + ∠CPA = 180**°**

∠BPD = ∠CPD

⇒ ∠BPA = ∠CPA

Therefore,

∆CAP ≅ ∆BAP **(ASA criteria)**

Hence, CP = BP.

**4. In the figure, BC = CE and ****∠****1 = ****∠****2. Prove that ∆GCB ****≅** **∆DCE.**

**Answer**

In ∆GCB and ∆DCE and

∠1 + ∠GBC = ∠2 + ∠DEC = 180**°**

∠1 = ∠2

⇒ ∠GBC = ∠DEC

BC = CE

∠GCB = ∠DCE ** (vertically opposite angles)**

Therefore,

∆GCB ≅ ∆DCE **(ASA criteria)**

**5. In ∆ABC, AB = AC and the bisectors of angles B and C intersect at point O. Prove that BO = CO and the ray AO is the bisector of angle BAC. **

**Answer**

In ∆ABC,

Since AB = AC

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

BO and CO are angle bisector of ∠B and ∠C respectively

Hence, ∠ABO = ∠OBC = ∠BCO = ∠ACO

Join AO to meet BC at D

In ∆ABO and ∆ACO and

AO = AO

AB = AC

∠C = ∠B

Therefore, ∆ABO ≅ ∆ACO **(SAS criteria)**

Hence, ∠BAO = ∠CAO

⇒ AO bisects angle BAC

In ∆ABO and ∆ACO

And AB = AC

AO = AO

∠BAD = ∠CAD ** (proved)**

∆ABO ≅ ∆ACO **(SAS criteria)**

Therefore,

BO = CO

**6. In the figure, AB = EF, BC = DE, AB and FE are perpendiculars on BE. Prove that ****∆ABD ****≅** **∆FEC**

**Answer**

In ∆ABD and ∆FEC

AB = FE

BD = CE **(BC = DE; CD is common)**

∠B = ∠E

∆ABD ≅ ∆FEC ** (SAS criteria)**

**7. In the figure, BM and DN are both perpendiculars on AC and BM = DN. Prove that AC bisects BD. **

**Answer**

In ∆BMR and ∆DNR

BM = DN

∠BMR = ∠DNR = 90**°**

∠BRM = ∠DRN **(vertically opposite angles)**

Hence, ∠MBR = ∠NDR **(sum of angles of a triangle = 180°)**

∆BMR ≅ ∆DNR **(ASA criteria)**

Therefore, BR = DR

So, AC bisects BD.

**8. In ∆PQR, LM = MN, QM = MR and ML and MN are perpendiculars on PQ and PR respectively. Prove that PQ = PR. **

**Answer**

In ∆QLM and ∆RNM

QM = MR

LM = MN

∠QLM = ∠RNM = 90**°**

Therefore,** **∆QLM ≅ ∆RNM **(RHS criteria)**

Hence, QL = RN **…(i)**

Join PM

In ∆PLM and ∆PNM and

PM = PM **(common)**

LM = MN

∠PLM = ∠PNM = 90**°**

Therefore, ∆PLM ≅ ∆PNM **(RHS criteria)**

Hence, PL = PN **…(ii)**

From (i) and (ii)

PQ = PR

**9. In the figure, RT = TS, ****∠****1 = 2****∠****2 and ****∠****4 = 2****∠****3. Prove that ∆RBT ****≅**** SAT. **

**Answer**

∠1 = 2∠2 and ∠4 = 2∠3

1 = 22 and 4 = 23∠1 = ∠4 **(vertically opposite angles)**

⇒ 2∠2 = 2∠3 or ∠2 = ∠3 **…(i)**

∠R = ∠S **(since RT = TS and angle opposite to equal sides are equal)**

⇒ ∠TRB = ∠TSA **…(ii)**

In ∆RBT and ∆SAT

RT = TS

∠TRB = ∠TSA

∠RTB = ∠STA **(common)**

Therefore, ∆RBT ≅ ∆SAT **(ASA criteria)**

**10. AD and BE are altitudes of an isosceles triangle ABC with AC = BC. Prove that AE = BD. **

**Answer**

In ∆CAD and ∆CBE

CA = CB **(Isosceles triangle)**

∠CDA = ∠CEB = 90**°**

∠ACD = ∠BCE **(common)**

Therefore, ∆CAD ≅ ∆CBE **(AAS criteria)**

Hence, CE = CD

But, CA = CB

⇒ AE + CE = BD + CD

⇒ AE = BD

**11. In ****∆ABC, X and Y are two points on AB and AC such that AX = AY. If AB = AC, prove that CX = BY. **

**Answer**

In ∆ABC

AB = AC

AX = AY

⇒ BX = CY

In ∆BXC and ∆CYB

BX = CY

BC = BC

∠B = ∠C = C **(AB = AC and angles opposite to equal sides are equal)**

Therefore, ∆BXC ≅ ∆CYB **(SAS criteria)**

Hence, CX = BY

**12. In the figure, AC = AE, AB = AD and ****∠****BAD = ****∠****EAC. Prove that BC = DE.**

**Answer**

In ∆ADE and ∆BAC

AE = AC

AB = AD

∠BAD = ∠EAC

∠DAC = ∠DAC = DAC **(common)**

⇒ ∠BAC = ∠EAD = EAD

Therefore, ∆ADE ≅ ∆BAC **(SAS criteria)**

Hence, BC = DE

**13. In the figure, ****∠****BCD = ****∠****ADC and ****∠****ACB = ****∠****BDA. Prove that AD = BC and ****∠****A = ****∠****B.**

**Answer**

∠BCD = ∠ADC

∠ACB = ∠BDA

∠BCD + ∠ACB = ∠ADC + BDA

⇒ ∠ACD = ∠BDCACD = BDC

In ∆ACD and ∆BCD

∠ACD = ∠BDCACD = BDC

∠ADC = ∠BCD

ADC = BCD CD = CD

Therefore, ∆ACD ≅ ∆BCD **(ASA criteria)**

Hence, AD = BC and ∠A = ∠B.

**14. In the figure, AP and BQ are perpendiculars to the line segment AB and AP = BQ. Prove that O is mid-point of the line segments AB and PQ.**

**Answer**

Since AP and BQ are perpendiculars to the line segment AB, therefore AP and BQ are parallel to each other.

In ∆AOP and ∆BOQ

∠PAO = ∠QBO = 90**°**

∠APO = ∠BQO **(alternate angles)**

AP = BQ

Therefore, ∆AOP ≅ ∆BOQAOP BOQ **(ASA criteria)**

Hence, AO = OB and PO = OQ

Thus, O is the mid-point of line segments AB and PQ.

**15. ∆ABC is isosceles with AB = AC. BD and CE are two medians of the triangle. Prove that BD = CE. **

**Answer**

CE is median to AB

⇒ AE = BE **…(i)**

BD is median to AC

⇒ AD = DC **…(ii)**

But AB = AC **…(iii)**

Therefore, from (i), (ii) and (iii)

BE = CD

In ∆BEC and ∆BDC

BE = CD

∠EBC = ∠DCB **(angles opposite to equal sides are equal)**

BC = BC **(common)**

Therefore, ∆BEC ≅ ∆BDC **(SAS criteria)**

Hence, BD = CE.

**16. Sides AB, BC and the median AD of ∆ABC are equal to the two sides PQ, QR and the median PM of ∆PQR. Prove that ∆ABC ****≅** **∆PQR. **

**Answer**

In ∆ABC and ∆PQR

BC = QR

AD and PM are medians of BC and QR respectively

⇒ BD = DC = QM = MR

In ∆ABD and ∆PQM

AB = PQ

AD = PM

BD = QM

Therefore, ∆ABD ≅ ∆PQMABD PQM **(SSS criteria)**

Hence, ∠B = ∠Q

Now in ∆ABC and ∆PQR

AB = PQ

BC = QR

∠B = ∠Q

Therefore, ∆ABC ≅ ∆PQRABC PQR **(SAS criteria)**

**17. Prove that in an isosceles triangle the altitude from the vertex will bisect the base. **

**Answer**

AB = AC

AD = AD

∠B = ∠C

Therefore, ∆ABD ≅ ∆ADC **(SSA criteria)**

Hence, BD = DC

Thus, AD bisects BC

**18. In ∆ABC, AB = AC. D is a point in the interior of the triangle such that ****∠****DBC = ****∠****DCB. Prove that AD bisects ****∠****BAC of ∆ABC. **

**Answer**

Since AB = AC

∠ABC = ∠ACB

But ∠DBC = ∠DCB

⇒ ∠ABD = ∠ACD

Now in ∆ABD and ∆ADC

AB = AC

AD = AD

∠ABD = ∠ACD

Therefore, ∆ABD ≅ ∆ADC **(SSA criteria)**

Hence, ∠BAD = ∠CAD

Thus, AD bisects ∠BAC.

**19. O is any point in the ∆ABC such that the perpendicular drawn from O on AB and AC are equal. Prove that OA is the bisector of ****∠****BAC. **

**Answer**

In ∆POA and ∆QOA

∠OPA = ∠OQA = 90**°**

OP = OQ** (given)**

AO = AO

Therefore,** **∆POA ≅ ∆QOA **(SSA criteria)**

Hence, ∠PAO = ∠QAO

Thus, OA bisects ∠BAC

**20. In ∆ABC, AB = AC, BM and CN are perpendicular on AC and AB respectively. Prove that BM = CN. **

**Answer**

∠BNC = ∠CMB = 90**°**

∠NBC = ∠MCB **(AB = AC)**

BC = BC

Therefore, ∆BNC ≅ ∆CMB** (AAS criteria)**

Hence, BM = CN

**21. ∆ABC is an isosceles triangle with AB = AC. GB and HC are perpendiculars drawn on BC. **

**Prove that**

**(i) BG = CH **

**(ii) AG = AH**

**Answer**

In ∆ABC

AB = AC

∠ABC = ∠ACB **(equal sides have equal angles opposite to them) ...(i)**

∠GBC = ∠HCB = 90**° ****…(ii) **

Subtracting (i) from (ii)

∠GBA = ∠HCA **…(iii)**

In ∆GBA and ∆HCA

∠GBA = ∠HCA **(from iii)**

∠BAG = ∠CAH **(vertically opposite angles)**

BC = BC

Therefore, ∆GBA ≅ ∆HCA **(ASA criteria)**

Hence, BG = CH and AG = AH

**22. In ∆ABC, AD is a median. The perpendiculars from B and C meet the line AD produced at X and Y. Prove that BX = CY. **

**Answer**

∠BXD = ∠CYD **(90°)**

∠XDB = ∠YDC **(vertically opposite angles)**

BD = DC **(AD is median on BC)**

Therefore, ∆BXD ≅ ∆CYD **(AAS criteria)**

Hence, BX = CY

**23. Two right-angled triangles ABC and ADC have the same base AC. If BC = DC, prove that bisects ****∠BCD.**

**Answer**

∠BAC = ∠DAC **(90°)**

BC = DC

AC = AC **(common)**

Therefore, ∆ABC ≅ ∆ADC **(SSA criteria)**

Hence, ∠BCA = ∠DCA

Thus, AC bisects ∠BCD

**24. PQRS is a quadrilateral and T and U are points on PS and RS respectively such that PQ = RQ, ****∠****PQT = ****∠****RQU and ****∠****TQS = ****∠****UQS. Prove that QT = QU. **

**Answer**

∠PQT = ∠RQU **…(i)**

∠TQS = ∠UQS **…(ii)**

Adding (i) and (ii)

∠PQS = ∠RQS

In ∆PQS and ∆RQS

∠PQS = ∠RQS

PQ = RQ **(given)**

QS = QS **(common)**

Therefore, ∆PQS ≅ ∆RQS **(SAS criteria)**

Hence, ∠QPS = ∠QRS

Now in ∆PQT and ∆RQU

∠QPS = ∠QRS

PQ = RQ **(given)**

∠PQT = ∠RQU **(given)**

Therefore, ∆PQT ≅ ∆RQU **(ASA criteria)**

Hence, QT = QU.

**25. A is any point in the angle PQR such that the perpendiculars drawn from A on PQ and QR are equal. Prove that ****∠****AQP = AQR. **

**Answer**

AM ⊥ PQ and AN ⊥ QR

AM = AN

In ∆AQM and ∆AQN,

AM = AN **(given)**

AQ = AQ **(common)**

∠AMQ = ∠ANQ **(Each 90°)**

So, by RHS congruence, we have

∆AQM ≅ ∆AQN

⇒ ∠AQM = ∠AQN **(c.p.c.t)**

⇒ ∠AQP = ∠AQR

**26. In the given figure P is a midpoint of chord AB of the circle O. Prove that OP ⟂ AB. **

**Answer**

Given:

In the figure, O is centre of the circle and AB is chord.

P is the mid-point of AB

⇒ AP = PB

To prove: OP ⊥ AB

**Construction:**join OA and OB

Proof:

In ∆OAP and ∆OBP

OA = OB **[radii of the same circle]**

OP = OP **[common]**

AP = PB **[given]**

∴ By Side - Side –Side criterion of congruency,

∆OAP ≅ ∆OBP

The corresponding parts of the congruent triangles are congruent.

∴ ∠OPA = ∠OPB

But ∠OPA + ∠OPB = 180**°**** ****[linear pair]**

∴ ∠OPA = ∠OPB = 90**°**

Hence OP ⊥ AB.

**27. In a circle with centre O. If OM is perpendicular to PQ, prove that PM = QM. **

**Answer**

Given:

In the figure, O is centre of the circle and PQ is a chord.

OM ⊥ PQ

**To prove:** PM = QM

**Construction:**Join OP and OQ

**Proof:**

In right triangles ∆OPM and ∆OQM,

OP = OQ ** [radii of the same circle]**

OM = OM **[common]**

∴ By right Angle-Hypotenuse-Side criterion of congruency,

∆ OPM ≅ ∆OQM

The corresponding parts of the congruent triangles are congruent.

∴ PM = QM

**28. In a triangle ABC, if D is midpoint of BC; AD is produced upto E such as DE = AD, then prove that:**

**a. DABD and DECD are congruent.**

**b. AB = EC**

**c. AB is parallel to EC**

**Answer**

**Given:**

D is mid-point of BC

⇒ BD = DC

DE = AD

**To prove:**

a. ∆ABD ≅ ∆ECD

b. AB = EC

c. AB || EC

BD = DC **(given)**

∠ADB = ∠CDE **(vertically opposite angles)**

AD = DE **(given)**

∴ By Side-Angle-Side criterion of congruence,

∆ABD ≅ ∆ECD

b. The corresponding parts of the congruent triangles are congruent.

∴ AB = EC

c. Also, ∠DAB = ∠DEC **(c.p.c.t)**

∴ AB || EC **(∠DAB and ∠DEC are alternate angles)**

**29. If the perpendicular bisector of the sides of a triangle PQR meet at I, then prove that the line joining from P,Q,R to I are equal. **

**Answer**

Given :

In ∆PQR,

PA is the perpendicular bisector of QR ⇒ QA = RA

RC is the perpendicular bisector of PQ ⇒ PC = QC

QB is the perpendicular bisector of PR ⇒ PB = RB

PA, RC and QB meet at I.

Proof: In ∆QIA and ∆RIA

QA = RA ** [Given]**

∠QAI = ∠RAI **[Each = 90°]**

IA = IA **[common]**

∴ By Side-Angle-Side criterion of congruence,

∆QIA ≅ ∆RIA

The corresponding parts of the congruent triangles are congruent.

∴ IQ = IR **…(i)**

Similarly, in ∆RIB and ∆PIB

RB = PB **…[Given]**

∠RBI = ∠PBI **….[Each = 90°]**

IB = IB **….[common]**

∴ By Side-Angle-Side criterion of congruence,

∆RIB ≅ ∆PIB

The corresponding parts of the congruent triangles are congruent.

∴ IR = IP **….(ii)**

From (i) and (ii), we have

**30. In the given figure ABCD is a parallelogram, AB is produced to L and E is a midpoint of BC. Show that:**

**a. DDCE**

**≅**

**DLBE**

**b. AB = BL**

**c. DC = AL/2**

**Answer**

Given:

ABCD is a parallelogram, where BE = CE

To prove:

a. ∆DCE ≅ ∆LBE

b. AB = BL

c. DC = AL/2

∠DCE = ∠EBL ** [DC || AB, alternate angles]**

CE = BE **[given]**

∠DEC = ∠LEB **[vertically opposite angles]**

∴ By Angle-Side-Angle criterion of congruence,

∆DCE ≅ ∆LBE

The corresponding parts of the congruent triangles are congruent.

∴ DC = LB **…(1)**

b. DC = AB **…(2) [Opposite sides of a parallelogram]**

From (1) and (2),

AB = BL **….(3)**

c. AL = AB + BL

⇒ AL = AB + AB **[From (3)]**

⇒ AL = 2AB

⇒ AL = 2DC **[From (2)]**

**32. In the given figure, AB = DB and AC = DC. Find the values of x and y. **

**Answer**In ∆ABC and ∆DBC

AB = DB ** [given]**

AC = DC **[given]**

BC = BC ** [common]**

∴ By Side-Side-Side criterion of congruence,

∆ABC ≅ ∆DBC

∴ ∠ACB = ∠DCB **[c.p.c.t]**

⇒ y + 15**° **= 63°

⇒ y = 63° - 15°

⇒ y = 48**°**

Now, ∠ABC = ∠DBC **[c.p.c.t]**

⇒ 29° = 2x - 4°

⇒ 2x = 29° + 4°

⇒ 2x = 33**° **

⇒ x = 33**°/**2

⇒ x = 16.5° and y = 48°