Angle Formed by Cevians in Isosceles Triangle

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Angle Formed by Cevians in Isosceles Triangle

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perpendicular bisector

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Source: 1997 USA MOP

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#1 h Jun 24, 2008, 12:45 AM • 2 Y

Y by Adventure10, Mango247

Consider $\triangle ABC$ with $AB=AC$ , and points $M$ on $AB$ and $N$ on $AC$ .
If $P = BN \cap CM$ , show that $MN || BC \iff \angle APM = \angle APN$ .

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#2 h Jan 30, 2009, 10:03 PM • 2 Y

Y by Adventure10, Mango247

First part: if $MN\|BC$ , then $A\hat{P}N=A\hat{P}M$

$BCNM$ is an isosceles trapezium, so $PB=PC$ (to see why, draw perpendiculars to $BC$ from $M$ and $N$ and look for congruent triangles). We also have $AB=AC$ , so $AP$ is the perpendicular bisector of $BC$ , and hence the bisector of $B\hat{A}C$ . Now the result becomes obvious (just angle chase).

Second part: if $A\hat{P}N=A\hat{P}M$ , then $MN\|BC$

Reflect everybody across $AP$ . Clearly line $ \line{BN}$ reflects to line $ \line{CM}$. If $AP$ is not the bisector of $B\hat{A}C$ , then the reflection of $C$ will be in ray $ \ray{NB}$ and will be different from $B$ , a contradiction ( $AC'=AB$ , angle chase again). So $AP$ is the bisector of $B\hat{A}C$ . From this, we see thet $M$ is the reflection of $N$ , so $AP$ is perpendicular bisector of $MN$ . But it is already the perpendicular bisector of $BC$ , so we are done.

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