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Prove that:
(i) ∆ABD  ≡ ∆ACD
(ii) ∠B = ∠C
(iii) ∠ADB = ∠ADC
(iv) ∠ADB = 90°

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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°

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