Question

1. In the figure, AB = AC and BD = BC. Prove that angle ACD = 3 times angle ADC.
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Answer 1

Step-by-step explanation:

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Hope this is correct one for the figure

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Answer 2

$\large {\underline {\underline{\color{red}{\boxed{\colorbox{black}{ αղรաҽɾ :-}}}}}}$

${\large{\bf{\red{\underline{\underline{Answer :-}}}}}}$${\purple}.{\green{\text{In ∆ABC , AC = AB</p><p>So, ∠ACB = ∠ CBA = α</p><p>∠BAC = 180°-2α</p><p>In ∆CBD , BC = BD</p><p>So, ∠CBD = 180°-α</p><p>∠BCD = ∠BDC = α/2</p><p>Now,</p><p>∠ACD = ∠ACB + ∠BCD = α+α/2 = 3α/2</p><p>∠ADC = ∠BDC = α/2</p><p>SO, ∠ACD = 3∠ADC}}}}}$Ho gaya bro...