A circle touches the side BC of a triangle ABC at P and the extended sides AB and AC at Q and R respectively. Prove that AQ is equal to half the perimeter of △ABC
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Step-by-step explanation:
Lengths of Δ drawn from an external pt to a circle are equal.
⇒ AQ=AR, BQ=BP, CP=CR$$
Perimeter of ΔABC=AB+BC+CA
=AB+(BP+PC)+(AR−CR)
=(AB+BQ)+PC+(AQ−PC)
=AQ+AQ=2AQ
AQ= 1/2(perimeter of ΔABC).
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