ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC.
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.
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)
ΔABD ≅ ΔACD
Also,
or,
(ii)
In ΔABE and ΔACE,
AB = AC (Given, ΔABC is an isosceles triangles)
AE = AE (Common side)
ΔABE ≅ ΔACE
Also, BE = CE (CPCT) .....(2)
And,
(iii)
In ΔBED and ΔCED,
BD = CD (Given, ΔDBC is an isosceles triangles)
BE = CE [From (2)]
DE = DE (Common side)
ΔBED ≅ ΔCED
Also,
(iv)
And,
(v)
$\because \angle A E B+\angle A E C=180^{\circ} \quad($ Linear pair $)$
$\Rightarrow \angle A E B+\angle A E B=180^{\circ} \quad[$ From $(3)]$
$\Rightarrow 2 \angle A E B=180^{\circ}$
$\Rightarrow \angle A E B=\frac{180^{\circ}}{2}$
$\Rightarrow \angle A E B=90^{\circ} \quad \ldots .(5)$
From (2) and (5), we get
AE is the perpendicular bisector of BC.