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Given an obtuse triangle \(ABC\)  with \(\angle ABC\) obtuse, extend \(\overline{AB}\) past \(B\) to a point \(D\) such that \(\overline{CD}\) is perpendicular to \(\overline{AB}\). Let \(F\) be the point on line segment \(\overline{AC}\) such that \(\overline{BF}\) is perpendicular to \(\overline{AB}\), and extend \(\overline{BF}\) past \(F\) to a point \(E\) such that \(\overline{BE}\) is perpendicular to \(\overline{CE}\). Given that \(\angle ECF = \angle BCD\), show that \(\triangle ABC \sim \triangle BFC\).

 

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 Jan 16, 2020
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Here are the images for the problem.

laugh

 Jan 16, 2020
edited by Omi67  Jan 16, 2020
edited by Omi67  Jan 16, 2020
edited by Omi67  Jan 16, 2020

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