A circle touches the side CD of a triangle PCD at E and touches PC and PD produced at A and B as shown in the figure. show tht PA = 1/2 (perimeter of triangle PCD)
Answers
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
Answer:
CE=AC[TANGENT DRAWN FROM COMMON POINT C]....................(1)
DE=DB[tangents drawn from common point D]....................................(2)
PA=PB[tangents drawn fom common point P].......................................(3)
PERIMETER OF PCD=PC+PD+CD
perimeter of PCD=PC+PD+CE+DE [FROM(1)]
'' '' '' =PC+PD+AC+DB
'' '' '' =[PC+AC]+[PD+DB]
'' '' '' = PA + PB
'' '' ''= PA + PA [FROM(3)]
PERIMETER OF PCD=2PA
⇒1/2 PERIMETER OF PCD =PA
Step-by-step explanation: