Question

pls answer me this question....​

Answer 1

Proof:

in ΔABD and ΔADC,

AB = AC (given)

BD = DC(midpoint)

∠ABD = ∠ACD(isoceles triangle)

by SAS, thes two Δs are congruent

by CPCT,

∠bad = ∠cad

⇒AD bisects ∠A

Also by CPCT

∠ADB = ∠ADC

but BDC is a straight line

⇒∠ADB + ∠ADC = 180°

⇒∠ADB = ∠ADC = 90°

⇒ AD⊥BC

hence, proved

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