AB,BC,CA are sides of triangle ABC touch the circle with centre O and radius r at P,Q and R respectively. ( Circle is inside the Triangle and point P is on AB, Q on BC and R on AC )
Prove that -
(1) AB+CQ=AC+BQ.
(2) Area of ABC = 1/2 Perimeter of ABC * r
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From figure
AQ=AR , BP=BQ , CP=CR
AQ=AR
AB+BQ=AC+CR
AB+BP=AC+CP (BQ=BP AND CR=CP)
Perimeter of triangle ABC
AB+BC+CA
=AB+(BP+PC)+CA
=AB+BP(AC+PC)
=2AB+2BP
=2(AB+BP)
=2AQ
=1/2Perimeter of triangle ABC
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