A circle touches the side CD of triangle PCD at Q and touches PC and PD produced at A
and B respectively. If PA=5cm, find perimeter of triangle PCD.
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PA = PB ( tangents from an external point to a circle are equal ).
CA = CE ------ 1
DB = DE ------ 2 ( Tangents from an external point to a circle are equal )
PA + PB =(PC + CA) + (PD + DB)
2PA= (PC + CE) + (PD + DE).( From 1 and 2 )
= PC + CD + PD
= Perimeter of triangle PCD
PA = 1/2 Perimeter of triangle PCD .
HENCE PROVED
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