# [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.

Note by Xuming Liang
3 years ago

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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.

- 3 years ago

Nice solution. +1

- 3 years 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.

- 3 years 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.

- 3 years ago