[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:

ABCABC is a right triangle at BB. DD is a point inside ABCABC such that AD=ABAD=AB. EE is on ACAC satisfying DEBDDE\perp BD. Suppose cirucmcircle (CDE)(CDE) intersects BCBC at FF. Prove that BE=EFBE=EF.

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

Note by Xuming Liang
4 years, 1 month ago

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Solution 1:

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

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

Now, observe that AB=AD=AD AB = AD = AD' so A A is the circumcenter of BDD \triangle B D D'. This means that BDD=12BAD \angle BD' D = \frac{1}{2} \angle BAD . This motivates constructing MM as the midpoint of BD BD , so that we get BDD=BAM \angle BD'D = \angle BAM . Hence, it suffices to show that EDB+BDD=ACB+BAM \angle ED'B + \angle BD'D = \angle ACB + \angle BAM .

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

Xuming Liang - 4 years, 1 month ago

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Nice solution. +1

Surya Prakash - 4 years, 1 month ago

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Solution 2:

From the first paragraph of solution 1, we know it suffices to prove 180=EDC+EBC180^{\circ}=\angle EDC+\angle EBC.

Let ω\omega denote the circle centered at AA with radius ABAB. Suppose DEω=FDE\cap \omega =F distinct from DD, CDω=GCD\cap \omega =G distint from DD. Since EDBDED\perp BD, FBFB is the diamenter and thus F,A,BF,A,B are collineaar.

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

Through DD construct a line parallel to ACAC to intersect FB,FGFB,FG at J,IJ,I respectively. By property of parallels JD=IJJD=IJ will imply AE=AHAE=AH. To show this, let KK be the midpoint of GDGD, then we want to prove JKGIJK||GI.

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

Xuming Liang - 4 years, 1 month ago

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Solution 3(Simplest one yet): We want to prove 180EDC=EBC180-\angle EDC=\angle EBC. The two angles seem too far away to be related, we attempt to bring them closer by constructing XX on ACAC such that BXACBX\perp AC. Since AD2=AB2=AXACAD^2=AB^2=AX*AC, we know ADCAXD    ADC=AXDADC\sim AXD\implies \angle ADC=AXD. Clearly ADE=90ADB=90ABD=DBC\angle ADE=90-\angle ADB=90-\angle ABD=\angle DBC, Therefore the problem is equivalent to 180EDCADE=EBCDBC180-\angle EDC-\angle ADE=\angle EBC-\angle DBC or CXD=EBD\angle CXD=\angle \angle EBD, which is true because BDEXBDEX is cyclic.

Xuming Liang - 4 years, 1 month ago

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