# Changes

,  05:33, 14 August 2014
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[[File:fig3.png|200px]]

[[File:fig3.png|200px]]

==Interpretation of problem==

==Interpretation of problem==
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# In the quadrilateral ABCD sides BC , DC & QB are given .
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'''==Concepts used=='''
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#Tangents drawn from an external point  to a circle are equal.
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#In a quadrilateral, if all angles are equal  and  a pair of adjacent sides are equal then it is a square
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#In a circle, the radius drawn at the point of contact is perpendicular to the tangent
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'''==Algorithm=='''
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In the fig  BC=38 cm  and  BQ=27 cm <br>
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BQ=BR=27 cm  (because by concept 1)    <br>
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Therefore,  CR=BC-BR=38-27=11 cm  <br>
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CR=SC=11 cm                                                            (because by concept 1)  <br>
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DC=25 cm  <br>
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therefore ,    DS=DC-SC=25-11=14 cm  <br>
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DS=DP=14 cm                                                              (because by concept 1)  <br>
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Also {AD ortho DC} , {OP ortho AD} and {OS ortho DC}  <br>
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∠D=∠S=∠P=90˚      <br>
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⇒  ∠O=90˚ <br>
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therefore DSOP is a Square  <br>
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SO=OP=14 cm    <br>  hence Radius of given circle is 14 cm
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=Problem 5=

=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 =$\frac{1}{2}$ [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 =$\frac{1}{2}$ [perimeter of  △ABC].
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