[Solutions Posted]Geometry Proof Problem of the day #1

I was browsing through my past photos and dug up an ample amount of geometry proof problems I've solved in the past. Starting today, I will select and share one of those problems and post it on here every few days. I hope you guys will find these problems interesting and fun.

This one caught my eye as the first problem:

$ABC$ is a right triangle at $B$ . $D$ is a point inside $ABC$ such that $AD=AB$ . $E$ is on $AC$ satisfying $DE\perp BD$ . Suppose cirucmcircle $(CDE)$ intersects $BC$ at $F$ . Prove that $BE=EF$ .

Looking forward to solutions.involving different methods. Exercise your creativity on the limitless space of geo.

#Geometry

Easy Math Editor

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Comments

Solution 1:

We want to show that $\angle EBF = \angle EFB$ . Since $\angle EFB = 180 ^ \circ - \angle EFC = 180^ \circ - \angle EDC$ , it suffices to show that $\angle EBC + \angle EDC = 180 ^ \circ$ .

We almost want to say that they are opposite angles of a cyclic quad, but $B, D$ lie on the same side of $EC$ , and so that won't be possible. This motivates the construction of $D'$ as the reflection of $D$ in the line $EC$ , and it suffices to show that $BED'C$ is a cyclic quad. There are many different approaches that we could use, and we will show that $\angle ECB = \angle ED'B$ . We pick this angle because $\angle ACB$ is fixed in this question, while $D, E, D'$ are variable points.

Now, observe that $AB = AD = AD'$ so $A$ is the circumcenter of $\triangle B D D'$ . This means that $\angle BD' D = \frac{1}{2} \angle BAD$ . This motivates constructing $M$ as the midpoint of $BD$ , so that we get $\angle BD'D = \angle BAM$ . Hence, it suffices to show that $\angle ED'B + \angle BD'D = \angle ACB + \angle BAM$ .

Observe that $AM \parallel ED$ since they are both perpendicular to $BD$ . We then have $\angle ACB + \angle BAM = 90 ^ \circ - \angle MAC = 90^ \circ - DEC = \angle EDD'$ . This is equal to $\angle ED'D$ due to isosceles triangle $ED'D$ . Hence we are done.

Xuming Liang - 5 years, 9 months ago

Nice solution. +1

Surya Prakash - 5 years, 9 months ago

Solution 2:

From the first paragraph of solution 1, we know it suffices to prove $180^{\circ}=\angle EDC+\angle EBC$ .

Let $\omega$ denote the circle centered at $A$ with radius $AB$ . Suppose $DE\cap \omega =F$ distinct from $D$ , $CD\cap \omega =G$ distint from $D$ . Since $ED\perp BD$ , $FB$ is the diamenter and thus $F,A,B$ are collineaar.

The problem is equivalent to $\angle EBC=\angle FBG$ or $\angle GBE=90^{\circ}$ . Since $FG\perp GB$ , we only have to prove $BE\parallel GF$ . Extend $AC$ to meet $GF$ at $H$ , it now suffices to show $AE=AH$ .(This is a well-known problem, proof is below)

Through $D$ construct a line parallel to $AC$ to intersect $FB,FG$ at $J,I$ respectively. By property of parallels $JD=IJ$ will imply $AE=AH$ . To show this, let $K$ be the midpoint of $GD$ , then we want to prove $JK||GI$ .

Because $\angle FGD=\angle FBD$ , the problem is now equivalent to $\angle JKD=\angle IGD=\angle FBD$ or $J,D,B,K$ are concyclic. Observe that $\angle AKC=90^{\circ}=\angle ABC$ since $K$ is the midpoint of chord $GD$ , hence $A,K,B,C$ are concyclic. This means $\angle CKB=\angle CAB=\angle DJB$ by parallels and is sufficient to establish $JDBK$ is cyclic.

Xuming Liang - 5 years, 9 months ago

Solution 3(Simplest one yet): We want to prove $180-\angle EDC=\angle EBC$ . The two angles seem too far away to be related, we attempt to bring them closer by constructing $X$ on $AC$ such that $BX\perp AC$ . Since $AD^2=AB^2=AX*AC$ , we know $ADC\sim AXD\implies \angle ADC=AXD$ . Clearly $\angle ADE=90-\angle ADB=90-\angle ABD=\angle DBC$ , Therefore the problem is equivalent to $180-\angle EDC-\angle ADE=\angle EBC-\angle DBC$ or $\angle CXD=\angle \angle EBD$ , which is true because $BDEX$ is cyclic.

Xuming Liang - 5 years, 9 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}`	$\boxed{123}$