Given: In the figure,
AD = AC
BD = CD
To prove :
(i) ∆ABD ≅ ∆ACD (ii) ∠B = ∠C
(iii) ∠ADB = ∠ADC (IV) ∠ADB = 90°
Proof : In ∆ABD and ∆ACD
AD = AD (common)
AD = AC (given)
BD = CD (given)
(i) ∴ ∆ABD ≅ ∆ACD (SSS axiom)
(ii) ∴ ∠B = ∠C (c.p.c.t.)
(iii) ∠ADB = ∠ADC (c.p.c.t)
But ∠ADB + ∠ADC = 180 (Linear pair)
∴ ∠ADB = ∠ADC
= 180°/2 = 90°