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Circles Tangents Problems (view source)
Revision as of 05:24, 14 August 2014
, 05:24, 14 August 2014→Problem 5
A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC].
A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =<math>\frac{1}{2}</math> [perimeter of △ABC].
[[File:123.png|300px]]
[[File:123.png|300px]]
==Algorithm==
In the figure AQ , AR and BC are tangents to the circle with center O.<br>
BP=BQ and PC=CR (Tangents drawn from external point are equal) ---------- (1)<br>
Perimeter of △ABC=AB+BC+CA<br>
=AB+(BP+PC)+CA<br>
=AB+BQ+CR+CA ------ (From eq-1)<br>
=(AB+BQ)+(CR+CA)<br>
=AQ+AR ----- (From fig)<br>
=AQ+AQ -- --- (∵∵∵∵∵∵∵∵∵∵∵∵∵∵∵∵∵∵AQ=AR)<br>
=2AQ<br>
∴ AQ = <math>\frac{1}{2}</math> [perimeter of △ABC]