Question

6. AABC and ADBC are two isondes tragjes
on the same base BC and vertices A and D
are on the same side of BC. AD is attended
to intersect BC at E, show that
Q) AABD=AACD
(m) AABELACE
mm) AE bisects LA as well as LD
(iv) AE is the perpendicular bisector of BC
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Answer 1

(i) In △ABD and △ACD,

AB = AC (since △ABC is isosceles)

AD = AD (common side)

BD = DC (since △BDC is isosceles)

Δ ABD ≅ Δ ACD (SSS test of congruence)

∴∠BAD = ∠CAD i.e. ∠BAP = ∠PAC --------------(1)

(ii) In △ABP and △ACP,

AB = AC (since △ABC is isosceles)

AP = AP (common side)

∠BAP = ∠PAC from (1)

△ABP ≅△ACP SAS test of congruence

∴ BP = PC -----------------(2)

∠APB=∠APC

(iii) Since △ABD ≅△ACD

∠BAD = ∠CAD from (1)

So, AD bisects ∠A

i.e. AP bisects ∠A. -------------(3)

In △BDP and △CDP,

DP = DP common side

BP = PC from (2)

BD = CD (since △BDC is isosceles)

△BDP ≅△CDP SSS test of congruence

∴∠BDP = ∠CDP

∴ DP bisects∠D

So, AP bisects ∠D -----------------(4)

From (3) and (4),

AP bisects ∠A as well as ∠D.

(iv) We know that

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

Also, ∠APB = ∠APC from (2)

∴ ∠APB = ∠APC = 180°/2 = 90∘

BP = PC and ∠APB = ∠APC = 90∘

Hence, AP is perpendicular bisector of BC.

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