Question

ΔABC and ΔDBC 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
(i) ΔABD ≅ ΔACD
(ii) ΔABE ≅ ΔACE
(iii) AE bisects ∠A as well as ∠D
(iv) AE is the perpendicular bisector of BC.

Answer 1

Given: ΔABC and ΔDBC are two isosceles triangles on the same base BC.

To prove:
(i) ΔABD ≅ ΔACD
(ii) ΔABE ≅ ΔACE
(iii) AE bisects ∠A as well as ∠D
(iv) AE is the perpendicular bisector of BC

Proof:
(i) In ΔABD and ΔACD,

BD = CD (Given, ΔDBC is an isosceles triangles)
AB = AC (Given, ΔABC is an isosceles triangles)
AD = AD (Common side)

$\therefore$ By SSS congruence criteria,
ΔABD ≅ ΔACD

Also, $\angle$BAD = $\angle$CAD (CPCT)
or, $\angle$BAE = $\angle$CAE .....(1)

(ii)
In ΔABE and ΔACE,

AB = AC (Given, ΔABC is an isosceles triangles)
$\angle$BAE = $\angle$CAE [From (i)]
AE = AE (Common side)

$\therefore$ By SAS congruence criteria,
ΔABE ≅ ΔACE

Also, BE = CE (CPCT) .....(2)
And, $\angle$AEB = $\angle$AEC (CPCT) .....(3)

(iii)
In ΔBED and ΔCED,

BD = CD (Given, ΔDBC is an isosceles triangles)
BE = CE [From (2)]
DE = DE (Common side)

$\therefore$ By SSS congruence criteria,
ΔBED ≅ ΔCED

Also, $\angle$BDE = $\angle$CDE (CPCT) .....(4)

(iv)
$\because$ $\angle$BAE = $\angle$CAE [From (1)]
And, $\angle$BDE = $\angle$CDE [From (4)]
$\therefore$ AE bisects ∠A as well as ∠D.

(v)
$$\begin{gathered} \because \angle AEB+\angle AEC=180°\left(\mathrm{Linear}\mathrm{pair}\right) \\ \Rightarrow \angle AEB+\angle AEB=180°\left[\mathrm{From}\left(3\right)\right] \\ \Rightarrow 2\angle AEB=180° \\ \Rightarrow \angle AEB=\frac{180°}{2} \\ \Rightarrow \angle AEB=90°.....\left(5\right) \end{gathered}$$

From (2) and (5), we get
AE is the perpendicular bisector of BC.