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.

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## Comments

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TopNewestSolution 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.

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Nice solution. +1

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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.

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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.

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