Question
The given figure shows a triangle $A B C$ in which $A D$ is perpendicular to side $B C$ and $B D=C D$. Prove that:
(i) $\triangle A B D \cong \triangle A C D$
(ii) $A B=A C$
(iii) $\angle B=\angle C$

Answer

(i) In the given figure Δ ABC
AD ⊥ BC, BD = CD
In Δ ABD and Δ ACD
AD = AD ............(common)
∠ADB = ∠ADC ...............(each 90°)
BD = CD ...........(Given)
∴ Δ ABD ≅ Δ CAD .........(By SAS Rule)
(ii) Side AB = AC .........(c.p.c.t.)
(iii) ∠B = ∠C
Reason, since Δ ADB ≅ Δ ADC
∴ ∠B = ∠C
Hence proved.

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