In the drawing below, $ABCD$ and $EFGC$ are squares. The lines $BG$ and $DE$ intersect at point $H$.

a) Verify that $\angle BHD=90^{\circ}$ and conclude that point $H$ is simultaneously on the circumferences of diameters $BD$ and $EG$.

b) Find the value of $\angle AHD + \angle DHG + \angle GHF$.

![](https://cdn.mathpix.com/cropped/2024_05_01_4a19aea8e08094ae376fg-12.jpg?height=463&width=727&top_left_y=683&top_left_x=546)
Solution

Before solving the problem, we need to make a comment about cyclic quadrilaterals. Consider a triangle $ABC$, its circumcircle $\Gamma$, and a point $P$ in the same half-plane as $C$ determined by the line $AB$. There are three possibilities for the point $P$: it can be outside $\Gamma$, on it, or inside it. Each of these situations is represented in the drawing below.
![](https://cdn.mathpix.com/cropped/2024_05_01_4a19aea8e08094ae376fg-12.jpg?height=446&width=1102&top_left_y=1617&top_left_x=357)

In the first situation, when $P$ is outside the circumcircle, consider the point $Q$ of intersection of one of the segments $PA$ or $PB$ with $\Gamma$. Since $\angle ACB$ and $\angle AQB$ are inscribed angles in the same arc $AB$, we have $\angle ACB = \angle AQB$. By the Exterior Angle Theorem, we can write

$$
\begin{aligned}
\angle ACB & = \angle AQB \\
& = \angle APB + \angle PAQ \\
& > \angle APB
\end{aligned}
$$

In the second situation, since $\angle ACB$ and $\angle APB$ are inscribed angles in the same arc $AB$, we have $\angle ACB = \angle APB$. Finally, in the third situation, consider the point of intersection of the lines $PA$ or $PB$ with $\Gamma$. Again by the Exterior Angle Theorem, we have

$$
\begin{aligned}
\angle ACB & = \angle AQB \\
& < \angle AQB + \angle PAQ \\
& = \angle APB
\end{aligned}
$$

Therefore, $\angle ACB = \angle APB$ if and only if the point $P$ is on the circumcircle of triangle $ABC$.

a) Since $EC = CG$, $DC = BC$, and $\angle BCG = \angle DCE = 90^\circ$, it follows that triangles $BCG$ and $ECD$ are congruent. Thus, $\angle CBG = \angle CDE$. By the previous observation, this guarantees that point $B$ is on the circumcircle of triangle $DCH$. However, we know that the circle passing through points $B$, $C$, and $D$ is the circle with diameter $BD$. Consequently, $A$, $B$, $C$, $D$, and $H$ are on the same circle with diameter $BD$ and $\angle BHD = \angle BCD = 90^\circ$. Moreover, since $\angle EHG = 180^\circ - \angle BHE = 90^\circ = \angle EFG$, it follows from the initial observation that $E$, $H$, $F$, and $G$ are on the same circle.

![](https://cdn.mathpix.com/cropped/2024_05_01_4a19aea8e08094ae376fg-13.jpg?height=706&width=925&top_left_y=1207&top_left_x=645)

b) Considering the circumcircle of quadrilateral $ABCD$, since $\angle AHD$ and $\angle ACD$ are inscribed in the same arc $AD$, it follows that $\angle AHD = \angle ACD$. Similarly, considering the circumcircle of square $EFGC$, we have $\angle GHF = \angle FCG$. Therefore,

$$
\begin{aligned}
\angle AHD + \angle DHG + \angle GHF & = \\
\angle ACD + \angle DHG + \angle FCG & = \\
45^\circ + 90^\circ + 45^\circ & = 180^\circ
\end{aligned}
$$

Thus, we can conclude that points $A$, $H$, and $F$ are collinear.