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# 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
4 years 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 $$\angle AED=2\angle BEC$$.

- 4 years ago

Thanks.

- 4 years 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 :)

- 4 years 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.

- 4 years ago

Thanks.

- 4 years ago