Question

1. A ABC and A DBC are two isosceles triangles on
the same base BC and vertices A and D are on the
same side of BC (see Fig. 7.39). IfAD is extended
to intersect BC at P. show that
A ABD=AACD
(0) ААВРААСР
(ii) AP bisects Z A as well as Z D.
(iv) AP is the perpendicular bisector of BC.
B
PB
Fig. 7.391​

Answer 1

Solution :-

(i) In ∆ABD and ∆ACD we have,

• AB = AC (∆ABC is isosceles ∆) .
• AD = AD (common side) .
• BD = DC (∆BDC is isosceles ∆) .

So,

• ΔABD ≅ ΔACD (By SSS congruence) .

therefore,

• ∠BAD = ∠CAD (By CPCT) .

Hence

• ∠BAP = ∠PAC --------------- Eqn.(1)

(ii) In ∆ABP and ∆ACP we have,

• AB = AC (∆ABC is isosceles ∆) .
• AP = AP (common side) .
• ∠BAP = ∠PAC {from Eqn.(1) }.

therefore,

• ∆ABP ≅ ∆ACP (By SAS congruence) .

Hence,

• BP = PC (By CPCT) --------------- Eqn.(2)
• ∠APB = ∠APC (By CPCT) .

(iii) As proved above,

• ∆ABD ≅ ∆ACD .

and,

• ∠BAD = ∠CAD

So,

• AD bisects ∠A.

i.e.

• AP bisects ∠A --------------- Eqn.(3)

In ∆BDP and ∆CDP now, we have,

• DP = DP (common side) .
• BP = PC (from Eqn.(2) .
• BD = CD (∆BDC is isosceles ∆) .

therefore,

• △BDP≅△CDP (By SSS congruence .)
• ∠BDP = ∠CDP (By CPCT.)

Hence,

• DP bisects ∠D .

Or,

• AP bisects ∠D --------------- Eqn.(4)

From Eqn.(3) and Eqn.(4) we get,

• AP bisects ∠A as well as ∠D.

(iv)

→ ∠APB + ∠APC = 180° (Linear pair) .

Also,

→ ∠APB = ∠APC {By Eqn.(2)} .

So,

→ ∠APB = ∠APC = (180/2) = 90° .

also,

• BP = PC .

and,

• ∠APB = ∠APC = 90° .

Hence,

• AP is perpendicular bisector of BC .

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