In a triangle $ABC$ a point $I$ is a centre of inscribed circle. A line $AI$ meets a side $BC$ in a point $D$ . A bisector of $AD$ meets lines $BI$ and $CI$ respectively in a points $P$ and $Q$. Prove that heights of triangle $PQD$ meet in the point $I$.

I've tried to show that sides of triangle $PQD$ are parallel to sides of triangle $ABC$ but it didn't work out. That's why I ask you for help.

  • $\begingroup$ try something else then. $\endgroup$ – uniquesolution Oct 27 '15 at 22:33
  • $\begingroup$ I've also tried to do it in cartesian coordiantes system $\endgroup$ – Michael Oct 27 '15 at 22:37


It suffices to show that $CI \perp PD$. Note that since $AP = PD$ and $BI$ is bisector of $\angle ABD$, point $P$ lies on the circumcircle of $\triangle ABD$ (on the midpoint of the arc). From this you can compute $\angle PDC = \angle IDC - \angle ADP = \angle IDC - \angle ABI$ and show it is $90^{\circ} - \frac12 \angle C$, which is all you need.

  • 2
    $\begingroup$ I think you meant $\angle IDC-\angle ABI$. $\endgroup$ – Element118 Oct 27 '15 at 23:40
  • $\begingroup$ Yep -- fixed, thanks. :) $\endgroup$ – Evan Chen Oct 28 '15 at 1:12
  • $\begingroup$ How to show that point $P$ lies on the circumcircle of the triangle $ABD$? $\endgroup$ – Michael Oct 28 '15 at 15:12
  • $\begingroup$ It's a standard lemma; see e.g. mit.edu/~evanchen/handouts/Fact5/Fact5.pdf $\endgroup$ – Evan Chen Oct 28 '15 at 22:48
  • $\begingroup$ I went through it but this lemma assumes that point $P$ (in lemma point $L$) lies on the circumcircle of the triangle $ABD$ (in lemma $ABC$). In this task we want to prove that this point will lie on the circumcircle of the triangle $ABD$. $\endgroup$ – Michael Oct 31 '15 at 15:49

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