I can still solve synthetic geo in my head?

by math_explorer, Dec 16, 2016, 9:06 PM

By angle chasing / a standard property of spiral similarities (Lemma 3 of http://yufeizhao.com/olympiad/geolemmas.pdf ), $\triangle PBF \sim \triangle PCE$, so \[ \frac{PB}{PC} = \frac{BF}{CE} = \frac{BD}{CD}, \]so by the converse Angle Bisector theorem $PD$ bisects $\angle BPC$. Since $PQ$ was defined to bisect $\angle BPC$, the conclusion follows.

(See if you can reconstruct the problem and diagram :) )

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Let $ABC$ be a triangle and let the incenter touch its sides at $D,E,F$. If $P$ is the intersection of $(AEF)$ and $(ABC)$ while $Q$ is the midpoint of arc $BC$ not containing $A$ on $(ABC)$, prove that $P,D,Q$ are collinear.

or something very close

by tastymath75025, Dec 17, 2016, 4:37 AM

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