[Solutions Posted]Geometry Proof Problem of the day #2 (diagram included)

It's been a day or two? I did not keep track. Found this one in an online conversation I had with a geometer friend of mine; it might be or based on an existing olympiad problem though. I never keep track anyway.

$\triangle ABC$ has a right angle at $A$ , $D\in BC$ and $AD\perp BC$ . $X$ is an arbitrary point of $AD$ . $E,F\in CX,BX$ respectively such that $AB=BE,AC=CF$ .

Suppose $BE\cap CF=G$ . Prove that $FG=GE$ .

Extra Credit: Let $XG\cap BC=Q$ . Prove that $E,F,D,Q$ are concyclic.

#Geometry

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Comments

Solution 1:

What is special about $E,F$ in relation to $D$ ? Well, we do know from right triangle $ABC$ that $BE^2=AB^2=BD\cdot DC, CF^2=AC^2=CD\cdot CB$ , implying $BDE\sim BEC, CDF\sim CFB$ .

Construct the orthocenter of $XBC$ denoted by $H$ . Then $\angle BED=\angle XCB=\angle BHD\implies B,H,D,E$ are concyclic. Since $BH\cap CX, \angle HEB=\angle HDB=90^{\circ}$ , we can easily deduce that $HE^2=HX\cdot HD$ (try extending $CX$ to meet $BH$ ). Likewise we can show $HF^2=HX\cdot HD$ , so $HE=HF$ . Since $\angle HEG=\angle HFG=90^{\circ}$ , therefore $GE=GF$ .

Extra Credit: Let $\odot DEF\cap BC=D,Q'$ , we will show that $Q'\in XG$ which implies $Q\equiv Q'$ .

Suppose $\odot DEF\cap BE,BF=E',F'$ . Then $\angle EE'Q'=\angle EDQ'=\angle XEB\implies XE||E'Q'$ . Analogously $Q'F'||FX$ . Furthurmore, $FG=GE$ implies that $FEF'E'$ is an isosceles trapezoid. Hence $EF||E'F'$ , establishing that $XEF, Q'E'F'$ are homothetic and thus $XQ'$ passes through the center of homothety which is $G$ . We are done.

Xuming Liang - 5 years, 9 months ago

Hey Xuming, please upload the solution. I am very eager to know about the proof. I tried alot but cant succeed. Atleast any HINT. :D

Surya Prakash - 5 years, 9 months ago

Wish granted! Next time you could tag my name to get my attention faster.

Xuming Liang - 5 years, 9 months ago

Nice one by revealing the hidden orthic configuration! :) :) :)

Nihar Mahajan - 5 years, 9 months ago

Thanks @Xuming Liang Nice solution.

Surya Prakash - 5 years, 9 months ago

The first part of this problem turns out is from IMO 2012.

Xuming Liang - 5 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$