Perpendiculars are drawn from angles A, B, C of an acute angled triangle...

Apr 2016
23
0
Lucknow, India
Question : Perpendiculars are drawn from vertices A, B, C of an acute angled triangle on the opposite sides and produced to meet the circumscribing circles. If these produced parts be p, q, r. Then show that
(a/p) + (b/q) + (c/r) = 2(tanA + tanB + tanC).
where a, b, c are the sides opposite to vertices A, B, C respectively. A, B, C are also the angles made at the respective vertices.

My attempt: I don't know how to calculate and represent p, q, r in the terms of a, b, c and A, B, C to proceed in this question. Please help.
 
Jun 2013
1,151
614
Lebanon
Label the points: perps and where they meet the circle

consider right triangles, it is not difficult to show that

\(\displaystyle \tan B + \tan C = \frac{a}{p}\)

etc.
 
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Apr 2016
23
0
Lucknow, India
Thanks, I figured it out. I was just not using the property that equal chords subtend equal angle at the circumference.
 
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