Jtdylktuy

Let $ABCD $ a paralelogram and $H $ the hortocenter of $triangle ABC $.

Let $PQ $ , $RS $ trough $H $ s.t. $PQ|| AB $ and $RS||BC $ and $Pin [DA], Rin [AB], Qin [BC] , Sin [CD] $.

I have to show that $P, R, Q, S $ are on a circle.

I tried to prove that $m (angle P)+ m (angle R)=180$.

If $angle ABC=frac{pi}{2}$, then $ABCD$ is a rectangle, and $H=B$. Clearly, we then have $P=A$, $S=C$, and $R=Q=B$, so the points $P$, $Q$, $R$, and $S$ do indeed lie on a circle. If $angle ACB=frac{pi}{2}$, then $H=C$, so $P=D$, $Q=S=C$, and $R=B$. Thus, $P$, $Q$, $R$, and $S$ are again concyclic. If $angle BAC=frac{pi}{2}$, then $H=A$, and so $P=R=A$, $Q=B$, and $S=D$. That is, $P$, $Q$, $R$, and $S$ lie on a single circle. From now on, we assume that $ABC$ is not a right triangle, so that $Hnotin{A,B,C}$.

Project $H$ (orthogonally) onto $BC$ and $AB$ at $X$ and $Y$, respectively. Note that the triangles $HAP$, $HXQ$, $HCS$, and $HXR$ are similar triangles. That is,
$$frac{HP}{HS}=frac{HA}{HC}text{ and }frac{HQ}{HR}=frac{HX}{HY},.$$
Thus,
$$frac{HPcdot HQ}{HScdot HR}=frac{HAcdot HX}{HCcdot HY},.$$
As $angle AXC=angle HXC=frac{pi}{2}=angle HYA=angle CYA$, the quadrilateral $AYXC$ is cyclic, so
$$HCcdot HX=HCcdot HY$$
by the Power-of-Point Theorem. This means
$$frac{HPcdot HQ}{HScdot HR}=frac{HAcdot HX}{HCcdot HY}=1,,$$
or $$HPcdot HQ=HScdot HR,.$$
By the converse of the Power-of-Point Theorem, we conclude that $PRQS$ is a cyclic quadrilateral.