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Question
Prove that:
- ∆ ABD ≅ ∆ ACD
- ∠B = ∠C
- ∠ADB = ∠ADC
- ∠ADB = 90°
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Solution
Given: In the figure,
AB = AC
BD = CD
To prove:
- Δ ABD ≅ Δ ACD
- ∠B = ∠C
- ∠ADB = ∠ADC
- ∠ADB = 90°
Proof: In Δ ABD and Δ ACD
AD = AD ...(common)
AB = AC ...(given)
BD = CD ...(given)
(i) ∴ Δ ABD ≅ Δ 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
(iv) ∠ADB = ∠ADC
= `(180°)/2`
= 90°
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