Changes
From Karnataka Open Educational Resources
592 bytes added
, 05:24, 14 August 2014
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| 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] |