Points $D$ and $E$ are taken on $[BC]$ and $[AC]$ of acute angled triangle $ABC$ such that $BD$ and $CE$ are angle bisectors. Projections of $D$ onto $BC$ and $BA$ are $P$ and $Q$, projections of $E$ onto $CA$ and $CB$ are $R$ and $S$. Let $AP \cap CQ=X$, $AS \cap BR=Y$ and $BX \cap CY=Z$. Show that $AZ \perp BC$.
Problem
Source: Turkey National Mathematical Olympiad 2021 P4
Tags: geometry, geometry proposed, angle bisector, Z -orthocenter
electrovector
06.01.2022 19:43
I will prove that $BX$ is perpendicular to $AC$, which will finish the question because if so, $Z$ will be the orthocenter because of the symmetry Let the perpendicular feet from $B$ to $AC$ be $T$. Proving $\frac{BP}{PC} \cdot \frac{CT}{TA} \cdot \frac{AQ}{QB}=1$ will be enough because of Ceva theorem. It is easy to see $BQ=BP$, hence we should prove $\frac{AQ}{PC} \cdot \frac{CT}{TA}=1$. Basic trigonometric manipulation gives $AQ=AD \cdot \text{cos} BAC$, $CP=CD \cdot \text{cos} BCA$, $CT=BC \cdot \text{cos} BCA$ and $TA=BA \cdot \text{cos} BAC$ and we are done since $\frac{AB}{BC}=\frac{AD}{DC}$
lazizbek42
06.02.2022 18:43
Let $K$ be the perpendicular base drawn from $B$ to $AC$. Lemma:$BK,CQ,AP$ lines concurrent. Proof:$\frac{AK}{CK} = \frac{AQ}{CP}$ can be found from a simple calculation.$\boxed{BP= BQ}$ by Ceva theorem $$\frac{BP}{CQ} \cdot \frac{CK}{AK} \cdot \frac{AQ}{BQ} = 1$$So $BX$ and $CY$ are altitudies $\implies AZ \perp BC$ $\blacksquare$
Jasurbek
06.02.2022 19:24
we use the cheva theorem The heights BX and CY and AZ intersect at one point.
GuvercinciHoca
09.02.2022 14:45
Let's call $\omega$ to the circle with diameter $BD.$ Let $M$ be the second point where $\omega$ cuts the line $AC.$ We are going to show that $BM,CQ,AP$ lines concurrent. So $Z$ will be the orthocenter of $\triangle ABC.$ Let $BA \cap PM=S.$ We have $MA\perp MB$ and $\angle SMA=\angle DMP=\angle QPD=\angle AMQ.$ Which means $MA$ bisects $\angle SMQ \implies (S,Q,A,B)=-1.$ So $BM,CQ,AP$ must concur. Done.
Nagibator007
31.10.2023 18:12
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