Let $ABC$ be a triangle and $D$ be a point on segment $AC$. The circumscribed circle of the triangle $BDC$ cuts $AB$ again at $E$ and the circumference circle of the triangle $ABD$ cuts $BC$ again at $F$. Prove that $AE = CF$ if and only if $BD$ is the interior bisector of $\angle ABC$.
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Tags: geometry, circumcircle
weaving2
18.12.2021 18:54
By power of point: $CF \cdot CB=CD \cdot CA$ $AE \cdot AB = AD \cdot AC$ $\implies \dfrac{CF}{AE} \cdot \dfrac{CB}{AB}=\dfrac{CD}{AD}$ Note that $\dfrac{CF}{AE}=1 \iff \dfrac{CB}{AB}=\dfrac{CD}{AD}\iff BD$ is the bisector of $\angle ABC$
dkshield
17.05.2023 05:24
It's a very easy problem. Let us first consider that: $BD$ is the interior bisector of $\angle ABC$ Be: $$\angle ABD = \angle DBC= \alpha$$ $\Rightarrow \angle FAD=\alpha, \angle ECD= \alpha$,also: $\angle DEC=\alpha, \angle FAD =\alpha$, so: $\angle ADE= \angle FDC = 2\alpha$ So: $AD=DF$, $DE =CD$ $$\Rightarrow \triangle ADE \cong \triangle FDC \Rightarrow AE = CF$$ We're done:
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dkshield
17.05.2023 05:35
Now, completing the proof: Let us first consider that: $AE=CF$ Be: $$\angle DBC = \beta, \angle ABD= \gamma$$$\Rightarrow \angle FAD=\beta, \angle ECD= \gamma$,also: $\angle DEC=\beta, \angle FAD =\beta$, so: $\angle ADE= \angle FDC = \beta + \gamma$ Be: $\angle BAF = \alpha$, $\Rightarrow \angle GFC = \alpha + \beta$ $$\Rightarrow \triangle ADE \cong \triangle FDC \Rightarrow ED=DC \Rightarrow \beta = \gamma$$We're done:
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