Complicated dependencies

by math_explorer, Dec 3, 2010, 10:41 AM

Problem:
In triangle $ABC$, $AD$ is a median and $BE$ is a cevian, and they intersect at $F$. Prove that $BE$ is the angle bisector of $\angle B$ iff $\frac{BF}{FE} = \frac{BC}{AB} + 1$.

Solution:
Suppose the third cevian passing through $F$ is $CP$. By Van Obel's Theorem, $\frac{BF}{FE} = \frac{BD}{DC} + \frac{BP}{PA}$. But $\frac{BP}{PA} = \frac{CE}{EA}$ by Ceva's Theorem, and obviously $\frac{BD}{DC} = 1$. The conclusion follows from the angle bisector theorem: $BE$ bisects $\angle B$ iff $\frac{CE}{EA} = \frac{BC}{AB}$.

Problem:
$ABCD$ is an inscribed quadrilateral. The diagonals intersect at $P$. $M, N$ are the midpoints of $AB$ and $CD$ respectively; $L, K$ are the feet of the altitudes from $P$ to $AD$ and $BC$, respectively. Prove that $MN \perp LK$.

Obvious fact:
The incircle-side tangency points and the excircle-side tangency points are symmetric about the midpoint.

Obvious fact:
$9c^6 - 14c^4 + 6c^2 = 0$ has only one real solution.

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