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As the picture shows, in $\triangle ABC$, $D$ is on $AB$ and $E$ is on $AC$, $AD=AE$, $DC$ and $EB$ interact at $F$, $FB=FC$.

Question: Whether it's always true that $AB=AC$? If it's always true, please explain it in steps, otherwise, give a counter example.

I tried to solve it, but found it hard to prove any conclusions.

picture

kabenyuk
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Jason Xue
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    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Jan 14 '24 at 13:14

1 Answers1

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A very simple proof:

Noticing that the problem can be seen as this: There is an isosceles triangle BFC, and we extend BF to point E and CF to point D, we want to prove that AD=AE is always equivalent to AB=AC. And what controls the lengths of all of them is FD and FE.

If DF=EF, it's easy to prove that AB=AC, and vice versa. So AB=AC is equivalent to DF=EF. Now we want to prove that DF=EF is equivalent to AD=AE.

Without loss of generality, let's assume DF>EF. Connect DE, we must have $\angle DEF>\angle EDF$.

Next, let's consider $\angle AEB$ and $\angle ADC$. $\angle AEB=\angle ACB+\angle EBC$, $\angle ADC=\angle ABC+\angle BCD$. Because $\angle EBC=\angle BCD$, their difference $\angle ADC-\angle AEB= \angle ABC-\angle ACB$, which must be positive, because we can choose a point $G$ on $DF$ such that $GF=EF$ and $\angle GBC=\angle ECB$. This shows that $\angle ABC>\angle ACB$. So we have $\angle ADC>\angle AEB$.

Combined with $\angle DEF>\angle EDF$, we get $\angle ADE>\angle AED$, $AE>AD$. Similarly, $AE<AD$ if $DF<EF$. So $AE=AD$ only if $DF=EF$. The proof that $DF=EF$ is equivalent to $AD=AE$ is complete.

Yi Jiang
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