Problem 1329

## Solution.

Let $\alpha, \beta$, and $\gamma$ denote the measures of the angles of triangle $ABC$ at vertices $A, B$, and $C$, respectively.

Since angles $\text{B E C}$ and $\text{B F C}$ are right angles, points $E$ and $F$ lie on the circle with diameter $\overline{B C}$, and quadrilateral $B C E F$ is cyclic. From this, it follows that $\text{E F A=180 - B F E= B C E=}$.

Similarly, since angles $\text{A E H}$ and $\text{A F H}$ are right angles, quadrilateral $A E H F$ is cyclic, and $P$ is the center of its circumcircle. Therefore, from the relationship between the inscribed and central angles over the chord $\overline{A E}$, we get $\text{A P E=2 A F E=2}$. From the isosceles triangle $A P E$, we conclude that $\text{P E A= 1 2 (180 - A P E )=90 -}$, and from triangle $A Q E$ we have

$\text{E Q A=180 - Q A B- Q E A=180 - Q A B- P E A=180 - - (90 - )=90 + -}$.

!

On the other hand, since angles $\text{B F H}$ and $\text{B D H}$ are right angles, it follows that quadrilateral $B D H F$ is also cyclic. From the relationship between the inscribed angles over the chords $\overline{D H}$ and $\overline{B F}$, we have

$$\text{R F H= A F H= D B H= C B E=90 - , R H F= B H F= B D F=180 - C D F= F A C= B A C=}$$

where in the second extended equality, we used the fact that quadrilateral $C D F A$ is also cyclic. From triangle $R F H$, we get

$$\text{F R H=180 -( R H F+ R F H)=180 - - (90 - )=90 + - .}$$

As a consequence in quadrilateral $R E Q F$, we have

$$\text{F R E+ E Q F= F R H+180 - E Q A=90 + - +180 - (90 + - )=180}$$

so it follows that this quadrilateral is also cyclic. From the equality of the inscribed angles over the chord $\overline{R F}$, we have $\text{F Q R= F E R}$, and from the fact that quadrilateral $H E A F$ is cyclic (and the relationship of the inscribed angles over the chord $\overline{F H}$), it follows that $\text{F E R= F E H= F A H= B A D=90 -}$.

Finally,

$$\text{F Q R= F A H=90 - ,}$$

from which it follows that lines $Q R$ and $A H$ are parallel. Since line $A H$ is perpendicular to $B C$, due to the previously mentioned parallelism, it follows that $Q R$ and $B C$ are also perpendicular, thus proving the statement.