# Geometry Again!

Following are some geometry questions I couldnot solve. They may be from other olympiads, and if their solutions are available online please point me to them.

1. For a pentagon $ABCDE,\ AE=ED,\ AB+CD=BC,$ and $\angle BAE + \angle CDE=180 ^{\circ}$. P.T. $\angle AED=2\angle BEC$.

2. Let P be a point inside a triangle with $\angle C=90 ^{\circ}$ such that $AP=AC$, and let M be the midpt of AB and CH be the altitude. Prove that PM bisects $\angle BPH$ iff $\angle A=60 ^{\circ}$. Note by Megh Parikh
5 years, 10 months ago

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For Problem 1. Consider the point $F$ in the ray $BA$ such that $AF=CD$ ($A$ between $B$ and $F$), then triangles $FAE$ and $CDE$ are congruent (using $AE=ED$ and $\angle BAE+\angle CDE=180^\circ$. Note that the triangles $CEB$ and $FEB$ are congruent because $EC=EF$ and $BC=AB+CD=AB+AF=BF$. Finally, $\angle BEC=\angle BEF=\angle BEA+\angle AEF=\angle BEA+\angle DEC$ therefore $\angle AED=2\angle BEC$.

- 5 years, 10 months ago

Thanks.

- 5 years, 10 months ago

This problem wouldn't be so hard once you notice a property that $P$ carries from $AC=AP$.

WLOG say that $H$ is between $A,M$. Since $AC=AH\times AB$, therefore $AP=AH*AB$ which is enough to show that $\triangle APH\sim \triangle ABP\Rightarrow \frac {PH}{PB}=\frac {AP}{AB}=\frac {AC}{AB}$. Since $PM$ bisects $\angle BPH\iff \frac {PH}{PB}=\frac {HM}{BM}$ combining the results above we continue the equivalence: $\iff \frac {AC}{AB}=\frac {HM}{BM}=\frac {2HM}{AB}$ $\iff AC=2HM$.

Instead to computing for $HM$, we can geometrically show that $AC=2HM\iff \angle A$. To relate $AC$ and $HM$, we introduce the midpoint of $AC$(I actually originally connected $GH$, but then I found a simpler way). Since $\angle AGM=\angle CHM=90$($CM=AM$) and $AG=\frac {AC}{2}=HM$, by $HL$ congruence we have $\angle GAM=\angle CMA=\angle ACM\Rightarrow \angle CAB=60$. Q.E.D

For the first one I have the same proof as Jorge :)

- 5 years, 10 months ago

As a crazy thought for this problem, we could also argue by considering when Apollonius' circle of $H,B;M$ and the circle centered at $A$ with radius $AC$ intersect.

- 5 years, 10 months ago

Thanks.

- 5 years, 10 months ago