Prove that aoc = 90°
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drop a perpendicular from o to. ab and cd where circle touches Ab and ca....
let the points be h and k on ab and ca respectively ...
Now.. ch=ck(tangents from external Point)
ch=r
so ch=ck=ok=oh
therefore all angles equal =90
therefore <aoc=2<koc=90
let the points be h and k on ab and ca respectively ...
Now.. ch=ck(tangents from external Point)
ch=r
so ch=ck=ok=oh
therefore all angles equal =90
therefore <aoc=2<koc=90
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