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Given a fixed triangle $ABC$. A circle pass through $B,C$ meets $AC,AB$ at $D,E$ respectively.$BD$ meet $CE$ at $F$. Let $H,G$ be the projections of $F$ on the internal and external bisector of angle $A$. Prove line $GH$ always passes through a fixed point enter image description here

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  • $\begingroup$ Where does this problem come from? Very nice and obviously very difficult $\endgroup$ – Oldboy Jun 29 '18 at 9:59
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Produce GF, AH, and GH to cut BC at X, Y, and Z respectively. Through Z, erect a perpendicular to cut GA produced at Q. It should be clear that (1) circle V (in violet, centered at V, diameter = XQ) passing through G, X, Z, Q; and (2) circle R (in red, centered at R, diameter = YQ) passing through A, Y, Z, Q can be formed. Note that $\angle XZQ = 90^0$.

enter image description here

Through E, (1) draw EJ // AC cutting circle $\omega$ at J; (2) draw EI // AY cutting circle $\omega$ at I.

After such construction, the bisected angles located at A originally are now translated to E. Then, $\alpha = \beta$. That is, $\triangle IBJ$ is isosceles.

Draw $IL \bot BJ$ cutting BJ at L. Produce IL to cut QK at O. Note that $\triangle OBI \cong \triangle OIJ$ and OBIJ is then a kite. Since B, I, and J are different points on the circumference of the same circle and they are equidistant from the point O, then O must the center of circle $\omega$.

Therefore, GH, when produced, will always cut BC at Z, the midpoint of the chord BC.

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  • $\begingroup$ +1 You stole my next assignment :) $\endgroup$ – Oldboy Jul 3 '18 at 15:42
  • $\begingroup$ @Oldboy You were given more than 6 days to hand-in yours. $\endgroup$ – Mick Jul 3 '18 at 16:07
  • $\begingroup$ Actually, I have spent those 6 days on your problem math.stackexchange.com/questions/2798092/… $\endgroup$ – Oldboy Jul 3 '18 at 16:11
  • $\begingroup$ @Oldboy I have spent more than $2 \times 6$ days on 2798092 and still got stuck at the very last step. I am at the stage of giving it up almost. That is why I diverted your attention to that harder problem so that I have time to handle an easier one. Sorry! $\endgroup$ – Mick Jul 3 '18 at 16:39

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