A Acute angled triangle

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Kazi_Zareer
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A Acute angled triangle

Unread post by Kazi_Zareer » Sat Aug 06, 2016 8:22 pm

$\triangle ABC$ is an acute angled triangle. Perpendiculars drawn from its vertices on the opposite sides are $AD$, $BE $ and $CF$. The line parallel to $ DF $ through $E$ meets $BC$ at $Y$ and $BA$ at $X.$ $DF$ and $CA $ meet at $Z$. Circumcircle of $XYZ$ meets $AC$ at $S.$ Given, $\angle B=33 $ degrees find the angle $\angle FSD$ with proof.
We cannot solve our problems with the same thinking we used when we create them.

samiul_samin
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Re: A Acute angled triangle

Unread post by samiul_samin » Fri Feb 02, 2018 2:10 am

This topic is already deiscussed and solved in this forum.

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