Question 1: In the adjoining figure, AB = AC and DB = DC . Prove that \angle ABD = \angle ACD .2018-08-11_8-38-23

Answer:

Consider \triangle ABC

Given AB = AC \Rightarrow \angle ABC = \angle ACB (angles opposite to equal sides of a triangle are equal)

Now consider \triangle BCD

Given DB = DC \Rightarrow \angle DBC = \angle DCB (angles opposite to equal sides of a triangle are equal)

Therefore \angle ABC - \angle DBC = \angle ACB - \angle DCB

\Rightarrow \angle ABD = \angle ACD . Hence proved.

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Question 2: In the adjoining figure, AB = AC . BE and CF are respectively the bisectors of \angle ABC and \angle ACB . Prove that \triangle EBC \cong \triangle FCB .2018-08-11_8-38-36

Answer:

Consider \triangle ABC

Given AB = AC \Rightarrow \angle ABC = \angle ACB (angles opposite to equal sides of a triangle are equal)

Therefore we can also say that  \angle FBC = \angle ECB

Since \angle ABC = \angle ACB

\Rightarrow \frac{1}{2} \angle ABC = \frac{1}{2} \angle ACB

\Rightarrow \angle EBC = \angle FCB

Now consider \triangle EBC and \triangle FCB

\angle EBC = \angle FCB

BC is common

\angle FBC = \angle ECB

Therefore \triangle EBC \cong \triangle FCB (by A.S.A theorem). Hence proved.

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Question 3: If  \triangle ABC is an isosceles triangle with AB=AC . Prove that perpendiculars from the vertices B and C to their opposite sides are equal.2018-08-11_8-38-52

Answer:

Consider \triangle ABC

Given AB = AC \Rightarrow \angle ABC = \angle ACB (angles opposite to equal sides of a triangle are equal)

Now consider \triangle EBC and \triangle DCB

\angle ABC = \angle ACB

BC is common

\angle BEC = \angle CDB (right angles)

Therefore \triangle EBC \cong \triangle DCB  (by A.S.A theorem). Hence proved.

\Rightarrow BD = CE

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Question 4: In the adjoining figure, it is given that \angle BAD = \angle BCE and AB = BC . Prove that \triangle ABD \cong \triangle CBE 2018-08-11_8-39-04

Answer:

Given \angle BAD = \angle BCE

and \angle EOA = \angle DOC (vertically opposite angles)

\Rightarrow \angle AEO = \angle CDO

\Rightarrow \angle CEB = \angle ADB

Now consider \triangle ABD and \triangle CBE

\angle BAD = \angle BCE

AB = AC (given)

\angle CEB = \angle ADB

\triangle ABD \cong \triangle CBE  (by A.S.A theorem). Hence proved.

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Question 5: In \triangle ABC, AB = AC , and the bisector of \angle ABC and \angle ACB intersect at O . Prove that BO = CO   and the \overrightarrow{OA} is the bisector of \angle BAC 2018-08-11_8-39-15

Answer:

Consider \triangle ABC

Given AB = AC \Rightarrow \angle ABC = \angle ACB (angles opposite to equal sides of a triangle are equal)

\Rightarrow \frac{1}{2} \angle ABC = \frac{1}{2} \angle ACB

\Rightarrow \angle OBC = \angle OCB

\Rightarrow \angle OBA = \angle OCA

\Rightarrow OB = OC (since sides opposite to equal angles are equal)

Now consider \triangle AOB and \triangle AOC

AB = AC (given)

\angle OBA = \angle OCA

OB = OC

Therefore \triangle AOB \cong \triangle AOC (by S.A.S theorem)

\Rightarrow \angle BAO = \angle CAO

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Question 6: In the adjoining figure, it is given AB = AC and BD = EC . Prove that \triangle ABE \cong \triangle CAD 2018-08-11_8-39-28

Answer:

Consider \triangle ABC

Given AB = AC \Rightarrow \angle ABC = \angle ACB (angles opposite to equal sides of a triangle are equal)

\Rightarrow \angle ABE = \angle ACD

We have BD = EC

\Rightarrow BD + DE = EC + DE

\Rightarrow BE = CD

Consider \triangle ABE and \triangle ACD

AB = AC (given)

\angle ABC = \angle ACB

BE = CD

Therefore \triangle ABE \cong \triangle ACD (by S.A.S. theorem)

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Question 7: In the adjoining figure, line l is the bisector of \angle QAP and B is any point on l . BP and BQ are perpendiculars from B to the arms of A . Show that: (i) \triangle APB \cong \triangle AQB (ii) BP = BQ or B is equidistant from the arms of \angle A .2018-08-11_8-39-48

Answer:

Since l is the bisector of \angle QAP

\Rightarrow \angle QAB = \angle PAB

\angle APB = \angle AQB (perpendicular lines)

AB is common.

Therefore \triangle AQB \cong \triangle APB (by A.S.A theorem)

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Question 8: In the adjoining figure AD is a median and BL, CM are perpendiculars drawn from B and C respectively on AD and AD produced. Prove that BL = CM 2018-08-11_8-40-02

Answer:

Consider \triangle BDL and \triangle CDM

\angle LDB = \angle CDM  (vertically opposite angles)

\angle BLD = \angle CMD (right angles)

BD = DC (AD is the median )

Therefore \triangle BDL \cong \triangle CDM (A.A.S theorem)

\Rightarrow BL = CM

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Question 9: In the adjoining figure, BM \perp AC and DN \perp AC . Also BM = DN . Prove that AC bisects BD 2018-08-11_8-40-19

Answer:

Consider \triangle BMR and \triangle DNR

\angle BMR = \angle DNR (right angles)

\angle BRM = \angle DRN (vertically opposite angles)

and BM = DN (given)

Therefor \triangle BMR \cong \triangle DNR (by A.A.S theorem)

\Rightarrow BR = DR . Hence proved.

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Question 10: In the adjoining figure, ABC is an isosceles triangle with AB = AC. BD and CE are two medians of the triangle. Prove that BD = CE . 2018-08-11_8-40-32

Answer:

Consider \triangle ABC

Given AB = AC \Rightarrow \angle ABC = \angle ACB (angles opposite to equal sides of a triangle are equal)

Also, AB = AC \Rightarrow \frac{1}{2} AB = \frac{1}{2} AC

\Rightarrow BE = DC

Consider \triangle BEC and \triangle CDB

BC is common

\angle ABC = \angle ACB

BE = DC

Therefore \triangle BEC \cong \triangle CDB (by S.A.S theorem)

Therefore BD = CE . Hence proved.

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Question 11: In the adjoining figure, PS = PR , \angle TPS = \angle QPR . Prove that PT = PQ 2018-08-11_8-40-44

Answer:

Consider \triangle PSR

Given PS = PR \Rightarrow \angle PSR = \angle PRS (angles opposite to equal sides of a triangle are equal)

\Rightarrow \angle PST = \angle PRQ

Consider \triangle PST and \triangle PRQ

\angle PST = \angle PRQ

PS = PR

\angle TPS = \angle QPR (given)

Therefore \triangle PST \cong \triangle PRQ

Therefore PT = PQ

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Question 12: In the adjoining figure, \triangle ABC and \triangle DBC are two triangles on the same base BC such that AB = BC and DB = DC . Prove that \angle ABD = \angle ACD 2018-08-11_8-40-59

Answer:

Consider \triangle ABC

Given AB = AC \Rightarrow \angle ABC = \angle ACB … … … … … (i) (angles opposite to equal sides of a triangle are equal)

Similarly in  \triangle DBC

Given DB = DC \Rightarrow \angle DBC = \angle DCB … … … … … (ii) (angles opposite to equal sides of a triangle are equal)

Subtracting (ii) from (i) we get

\angle ABC - \angle DBC = \angle ACB - \angle DCB

\angle ABD = \angle ACD

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Question 13:  In the adjoining figure,  AB \parallel CD . O is the midpoint of AD . Show that (i) \triangle AOB \cong \triangle DOC (ii) O is also the midpoint of BC 2018-08-11_8-41-07

Answer:

(i) Consider \triangle AOB and \triangle DOC

OA = OD (given)

Since AB \parallel CD

\Rightarrow \angle ABO = \angle DCO

also \angle AOB = \angle DOC (vertically opposite angles)

Therefore \triangle AOB \cong \triangle DOC (by A.A.S theorem)

(ii) Since \triangle AOB \cong \triangle DOC

\Rightarrow OC = OB

\Rightarrow O is the midpoint of BC

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Question 14: In a \triangle ABC , it is given that AB =AC and the bisectors of \angle B and \angle C intersect at O . If M is a point on BO produced, prove that \angle MOC = \angle ABC 2018-08-11_8-48-31.jpg

Answer:

Consider \triangle ABC

Given AB = AC \Rightarrow \angle ABC = \angle ACB (angles opposite to equal sides of a triangle are equal)

\Rightarrow \frac{1}{2} \angle ABC = \frac{1}{2} \angle ACB

\Rightarrow \angle OBC = \angle OCB

In \triangle OBC , we have

\angle MOC = \angle OBC + \angle OCB

\Rightarrow \angle MOC = \angle OBC + \angle OBC

\Rightarrow \angle MOC = 2 \angle OBC = \angle ABC . Hence proved.

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Question 15: P is a point on the bisector of an \angle ABC . If the line through P parallel to AB meets BC at Q , prove that the \triangle BPQ is isosceles. 2018-08-11_8-48-48

Answer:

Since AB \parallel PQ

\angle ABP = \angle BPQ (alternate angles)

Also since BP bisects \angle ABQ

\angle ABP = \angle QBP

Hence \angle QBP = \angle BPQ

\Rightarrow BQ = QP (sides opposite equal angles in a triangle are equal).

Hence \triangle BQP is an isosceles triangle.

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Question 16: ABC is a triangle in which \angle B = 2 \angle C . D is a point on BC such that AD bisects \angle BAC and AB = CD . Prove that \angle BAC = 72^o 2018-08-11_8-56-41

Answer:

In \triangle ABC, \angle ABC = 2 \angle ACB

Let \angle ABC = 2y , where \angle ACB = y

AD is the bisector of \angle BAC . So \angle BAD = \angle  CAD = x

Let BP be the bisector of \angle ABC .

In \triangle BPC , we have \angle CBP = \angle BCP = y \Rightarrow BP = PC

In \triangle ABP and \triangle DCP , we have

\angle ABP = \angle DCP = y

AP = DC (given) and BP = PC

Therefore \triangle ABP \cong \triangle DCP (by S.A.S theorem)

\Rightarrow \angle BAP = \angle CDP and AP = DP

\Rightarrow \angle  CDP = 2x and \angle  ADP = x

In \triangle ABD , we have 

\angle ADC = \angle  ABD + \angle  BAD

\Rightarrow x+2x = 2y + x \Rightarrow x = y

In \triangle ABC , we have

\angle  BAC + \angle ABC + \angle ACB = 180^o

\Rightarrow 2x+2y + y = 180^o

\Rightarrow 5x = 180^o  \Rightarrow x = 36^o

Hence \angle BAC = 2x = 72^o

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