Question

1.∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P. Show that:(i)∆ABD ≅∆ACD(ii)∆ABP ≅∆ACP(iii)AP bisects ∠A as well as ∠D.(iv)AP is the perpendicular bisector of BC.

Answer 1

Step-by-step explanation:

(i) In △ABD and △ACD,  

AB=AC         ....(since △ABC is isosceles)  

AD=AD        ....(common side)

BD=DC        ....(since △BDC is isosceles)  

ΔABD≅ΔACD     .....SSS test of congruence,  

∴∠BAD=∠CAD i.e. ∠BAP=∠PAC          .....[c.a.c.t]......(i)

(ii) In △ABP and △ACP,  

AB=AC          ...(since △ABC is isosceles)

AP=AP         ...(common side)

∠BAP=∠PAC      ....from (i)

△ABP≅△ACP    .... SAS test of congruence

∴BP=PC      ...[c.s.c.t].....(ii)

∠APB=∠APC     ....c.a.c.t.

(iii) Since △ABD≅△ACD

∠BAD=∠CAD     ....from (i)

So, AD bisects ∠A

i.e. AP bisects∠A.....(iii)

In △BDP and △CDP,  

DP=DP     ...common side

BP=PC     ...from (ii)

BD=CD    ...(since △BDC is isosceles)  

△BDP≅△CDP    ....SSS test of congruence  

∴∠BDP=∠CDP    ....c.a.c.t.

∴ DP bisects∠D  

So, AP bisects ∠D        ....(iv)

From (iii) and (iv),  

AP bisects ∠A as well as ∠D.

(iv) We know that  

∠APB+∠APC=180                                  ....(angles in linear pair)

Also, ∠APB=∠APC       ...from (ii)

∴∠APB=∠APC =  =90

°

BP=PC and ∠APB=∠APC=90°

Hence, AP is perpendicular bisector of BC.

     

Answer 2

Figure:-

Given:-

• ∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC.
• AD is extended to intersect BC at P.

To Find:-

• (i)∆ABD ≅∆ACD(ii)∆ABP ≅∆ACP(iii)AP bisects ∠A as well as ∠D.(iv)AP is the perpendicular bisector of BC.

Solutions:-

(i). In ∆ABD and ∆ACD,

AB = AC (Given)

BD = CD (Given)

AD = AD (Common)

.:. ∆ABD ~ ∆ACD (By SSS Congruence rule)

<BAD = <CAD (By CPCT)

<BAP = <CAP ...........(1).

(ii). In ∆ABP and ∆ACP,

AB = AC (Given)

<BAP = <CAP [from equation (i).]

AP = AP (Common)

.:. ∆ABP ~ ∆ACP (By SAS Congruence rule)

BP = CP (By CPCT).........(2).

(iii). From equation (1),

<BAP = <CAP

hence, AP bisects <A

In ∆BDP and ∆CDP,

BD = CD (Given)

DP = DP (Given)

DP = CP [From equation (2)]

.:. ∆BDP ~ ∆CDP (By SSS Congruence rule)

<BDP = <CDP (By CPCT).......(3).

hence, AP bisects <D.

(iv). ∆BDP ~ ∆CDP

<BDP = <CDP (By CPCT).......(4).

<BDP + <CDP = 180° (Linear pair angles)

<BDP + <CDP = 180°

2<BPD = 180° [From equation (4)]

<BPD = 90° (5).

from equation (2) and (5), it can be said that AP is the perpendicular bisector of BC.