Question

$\Delta ABCand\Delta DBC\text{are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at E, show that}\left(i\right)\Delta ABD\cong \Delta ACD\left(ii\right)\Delta ABE\cong \Delta ACE\left(iii\right)\text{AE bisects}\angle A\text{as well as}\angle D\left(iv\right)\text{AE is the perpendicular bisector of BC.}$

Answer 1

Given,

ΔABC and ΔDBC are two isosceles triangles in which AB = AC & BD = DC.

(i) In ΔABD and ΔACD,
AD = AD (Common)
AB = AC (given)
BD = CD (given)
Therefore, ΔABD ≅ ΔACD (by SSS congruence rule)
∠BAD = ∠CAD (CPCT)
∠BAE = ∠CAE


(ii) In ΔABE & ΔACE,
AE = AE (Common)
∠BAE = ∠CAE (Proved above)
AB = AC (given)

Therefore,
ΔABE ≅ ΔACE
(by SAS congruence rule).


(iii)
∠BAD = ∠CAD (proved in part i)

Hence, AE bisects ∠A.
also,

In ΔBED and ΔCED,
ED = ED (Common)
BD = CD (given)
BE = CE (ΔABP ≅ ΔACP so by CPCT.)

Therefore, ΔBED ≅ ΔCED (by SSS congruence rule.)

Thus,
∠BDE = ∠CDE( by CPCT.)

Hence, we can say that AE bisects ∠A as well as ∠D.


(iv)
∠BED = ∠CED
(by CPCT as ΔBED ≅ ΔCED)

& BE = CE (CPCT)
also,

∠BED + ∠CED = 180° (BC is a straight line.)
⇒ 2∠BED = 180°
⇒ ∠BED = 90°

Hence,
AP is the perpendicular bisector of BC.