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

  • $\begingroup$ Where does this problem come from? Very nice and obviously very difficult $\endgroup$ – Oldboy Jun 29 '18 at 9:59

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.

  • $\begingroup$ +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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.