A Acute angled triangle
- Kazi_Zareer
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$\triangle ABC$ 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.
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Re: A Acute angled triangle
This topic is already deiscussed and solved in this forum.