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Question-189898




Question Number 189898 by Mingma last updated on 23/Mar/23
Answered by HeferH last updated on 24/Mar/23
Commented by HeferH last updated on 24/Mar/23
HD = diameter⇒  ∠HGD = 90° ⇒   ∠HGD  + ∠HCD= 180° ⇒ cyclic quadrilateral
$$\mathrm{HD}\:=\:\mathrm{diameter}\Rightarrow \\ $$$$\angle\mathrm{HGD}\:=\:\mathrm{90}°\:\Rightarrow\: \\ $$$$\angle\mathrm{HGD}\:\:+\:\angle\mathrm{HCD}=\:\mathrm{180}°\:\Rightarrow\:\mathrm{cyclic}\:\mathrm{quadrilateral} \\ $$$$\: \\ $$
Commented by Rupesh123 last updated on 24/Mar/23
Excellent!

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