lasha wrote: Let $ ABCD$ be a convex quadrilateral. Points $ P,Q$ and $ R$ are the feets of the perpendiculars from point $ D$ to lines $ BC, CA$ and $ AB$, respectively. Prove that $ PQ = QR$ if and only if the bisectors of the angles $ ABC$ and $ ADC$ meet on segment $ AC$. ABCD should be cyclic quadrilateral it is the 4th problem from IMO 2003 My solution: Using the Simson's line theorem, we get that the points P, Q and R are collinear. By Menelaus Theorem (for triangle ABC which is intersected by R-Q-P) and similarity of triangles ARD and CPD we get that PQ=QR iff AB:BC=AD:DC, that is, PQ=QR iff the angle bisectors of ABD and ADC meet on AC. Q.E.D.

ABCD can be not cyclic and the statement remains valid.

Prove it

1).The bisectors of the angles $ \angle{ABC}$ and $ \angle{ADC}$ meet on segment in segment $ [AC]\Leftrightarrow |AB|.|CD| = |AD|.|BC|$ 2). If $ R\mbox{ - is circumradius of} \triangle {ABC}$, then: $ \begin{array}{c}|QR| = |AD|.\sin{A} = |AD|.\frac {|BC|}{R} \\ |QP| = |DA|.\sin{C} = |AD|.\frac {|AB|}{R} \\ |AB|.|CD| = |AD|.|BC|\end{array}\}\Rightarrow |QP| = |QR|.$

using the Simson's line we get that $P,Q,R$ are collinear, so quadrilateral $AQDP$ is cyclic so $\angle PDQ= \angle BAC= \angle BDC$ because $ABCD$ is cyclic, so if bisectors meet on $AC$ it means $\frac {AB}{BC}= \frac {AD}{CD}$ so it means $ABCD$ is harmonic so $BD$ is symmedian $\triangle ACD$ is simillar to $\triangle PRD$ and if $BD$ is symmedian $DQ$ is median because they they are isogonals and angles are same