At the points A,B,C tangents are drawn to the circumcircle of triangle ABC. These tangents enclose a triangle PQR. Prove that its angles and sides are respectively $180^{\circ} - 2 A, 180^{\circ} - 2 B, 180^{\circ} - 2 C$ and $\frac{a}{2 c o s B c o s C}, \frac{b}{2 c o s C c o s A}, \frac{c}{2 c o s A c o s B}$

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Solution

Angles of ABC are obvious from the figure as
$∠ p = 90^{\circ} - A + 90^{\circ} - A = 180^{\circ} - 2 A$
sides opposite vertex Q is
RP = BP + BR = R (tan A + tan C)
= R $\frac{s i n (A + C)}{c o s A c o s C} = \frac{R s i n B}{c o s A c o s C}$
$= \frac{b}{2 c o s A c o s C}$ etc .
$[∵ \frac{a}{s i n A} = 2 R e t c]$

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