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Let $ABC$ be a triangle, and $BD$ be the angle bisector of $\angle B$. Let $DF$ and $DE$ be altitudes of $\triangle ADB$ and $\triangle CDB$ respectively, and $BI$ is an altitude of $\triangle ABC$. Prove that $AE$, $CF$ and $BI$ are concurrent.

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I saw that $BEDF$ is a kite as well as a cyclic quadrilateral, and even $I$ would lie on that circle. My approach was to assume $H$ as the intersection of $AE$ and $CF$, and prove that $BH\perp AC$, but it didn't work. Can anyone help? :)

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  • $\begingroup$ $AI$ should not be altitude of $\Delta ABC$. It must be $BI$ $\endgroup$ – Hardey Pandya Jan 19 '15 at 15:49
  • $\begingroup$ 've you applied Ceva's theorem ? $\endgroup$ – Hardey Pandya Jan 19 '15 at 15:51
  • $\begingroup$ @user91374 Hmm, I am not very familiar with it, so I did not use it. Let me see again. $\endgroup$ – Sawarnik Jan 19 '15 at 16:02
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    $\begingroup$ Perhaps you can do it. If it won't done, then no problem - we have to find another method. $\endgroup$ – Hardey Pandya Jan 19 '15 at 16:04
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By the angle bisector theorem, we have: $$\frac{CE}{AF}=\frac{CD\cos C}{AD\cos A}=\frac{BC\cos C}{AB\cos A}=\frac{CI}{AI}$$

As $BF=BE$, by the converse of Ceva's theorem, we have that $AE, CF$ and $BI$ are concurrent.

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Consider that: $$\frac {AF}{BF}=\frac{FA}{FD}\cdot \frac{FD}{FB}=\cot A\cdot \tan \frac{B}{2}$$ $$\frac {BE}{EC}=\frac{BE}{DE}\cdot \frac{DE}{CE}=\cot \frac{B}{2}\cdot \tan C$$ $$\frac{IC}{IA}=\frac{IC}{IB}\cdot \frac{IB}{IA}=\cot C\cdot \tan A$$

Multiply those equalities side by side, $AE,BF,CI$ are concurrent by Ceva theorem.

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$$\frac{CE}{BE}\cdot\frac{BF}{FA}\cdot\frac{AI}{IC}= \frac{CD\cos C}{BD\cos B/2}\cdot \frac{BD\cos B/2}{AD\cos A}\cdot\frac{AB\cos A}{BC\cos C}=\frac{CD}{AD}\cdot\frac{AB}{BC}=\frac{BC}{AB}\cdot\frac{AB}{BC}=1$$

Applying mutual Ceva's theorem, statement is demonstrated.

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