Answer 1

Proof:
In Δ ADB and Δ ADC,
AB = AC .........(given)
∠1 = ∠2 ............(given)
AD = AD .............(common)
∴ Δ ADB ≅ Δ ADC ................(S.A.S. Axiom)

(i) Hence ∠B = ∠C ..............(c.p.c.t.)

(ii) BD = DC ...............(c.p.c.t.)

(iii) and ∠ADB = ∠ADC ............(c.p.c.t.)
But ∠ADB + ∠ADC = 180° ....(Linear pair)
∴ ∠ADB = ∠ADC = 90°
Hence AD is perpendicular to BC.

Hence proved.