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Question
In the given figure ;
∠1 = ∠2 and AB = AC.
Prove that:
(i) ∠B = ∠C
(ii) BD = DC
(iii) AD is perpendicular to BC.
Solution
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.
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