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.

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

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**\(\angle A=60 ^{\circ}\).

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TopNewestFor 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 \(\angle AED=2\angle BEC\). – Jorge Tipe · 3 years, 1 month ago

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– Megh Parikh · 3 years, 1 month ago

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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 :) – Xuming Liang · 3 years, 1 month ago

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– Xuming Liang · 3 years, 1 month 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.Log in to reply

– Megh Parikh · 3 years, 1 month ago

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