$CH$ is an altitude in a right triangle $ABC$ $(\angle C = 90^{\circ})$. Points $P$ and $Q$ lie on $AC$ and $BC$ respectively such that $HP \perp AC$ and $HQ \perp BC$. Let $M$ be an arbitrary point on $PQ$. A line passing through $M$ and perpendicular to $MH$ intersects lines $AC$ and $BC$ at points $R$ and $S$ respectively. Let $M_1$ be another point on $PQ$ distinct from $M$. Points $R_1$ and $S_1$ are determined similarly for $M_1$. Prove that the ratio $\frac{RR_1}{SS_1}$ is constant.
Problem
Source: Kazakhstan national olympiad 2022, grade 9, p1
Tags: geometry, ratio, Kazakhstan
User126784
07.05.2022 15:17
Let the lines $RS$ and $R_1$$S_1$ intersect at point $O$. Since $H$, $R$, $P$, $M$ lie on the same circle with diameter $HR$, and the points $P$, $H$, $R_1$, $M_1$ lie on the circle with diameter $HR_1$, then ∠($MR, HR$) = ∠($MP, HP$) = ∠($M_1$$R_1$, $HR_1$).Similarly, ∠($HS, MS$) = ∠($HQ$, $M_1$$Q$) = ∠($H$$S_1$, $MS_1$). Therefore, the triangles $H$$R_1$$R$ and $H$$S_1$$S$ are similar, and in the segments $H$$P$ and $H$$Q$ are the corresponding heights. Therefore $R$$R_1$/$S$$S_1$ = $H$$P$/$H$$Q$ = const
SolveForChocolate
18.10.2022 11:43
User126784 wrote: Let the lines $RS$ and $R_1$$S_1$ intersect at point $O$. Since $H$, $R$, $P$, $M$ lie on the same circle with diameter $HR$, and the points $P$, $H$, $R_1$, $M_1$ lie on the circle with diameter $HR_1$, then ∠($MR, HR$) = ∠($MP, HP$) = ∠($M_1$$R_1$, $HR_1$).Similarly, ∠($HS, MS$) = ∠($HQ$, $M_1$$Q$) = ∠($H$$S_1$, $MS_1$). Therefore, the triangles $H$$R_1$$R$ and $H$$S_1$$S$ are similar, and in the segments $H$$P$ and $H$$Q$ are the corresponding heights. Therefore $R$$R_1$/$S$$S_1$ = $H$$P$/$H$$Q$ = const Can you explain pls how you get that triangle $HSS_1$ is similar to triangle $HRR_1$