In the following drawing, the chords $D E$ and $B C$ are perpendicular, with $B C$ being a diameter of the circle with center at $A$. Additionally, $\angle C G F=40^{\circ}$ and $G H=2 \text{~cm}$.

a) Determine the value of the angle $\angle C H F$.
b) Find the length of $H J$.

![](https://cdn.mathpix.com/cropped/2024_05_01_82072e76ed187b592b1cg-32.jpg?height=753&width=663&top_left_y=575&top_left_x=583)
Solution

a) Since $BC$ is a diameter, it follows that $\angle BFC=90^{\circ}$. Thus, as we also have $\angle CHG=90^{\circ}$, the circle $\Gamma$ with diameter $CG$ passes through $F$ and $H$. In this circle, the angles $\angle CGF$ and $\angle CHF$ are inscribed in the same arc $CF$, so $\angle CHF = \angle CGF = 40^{\circ}$.

b) Observing the circle $\Gamma$ again, we can conclude that $\angle HCG = \angle HFG$, as both are inscribed in the arc $GH$. Considering now the circle with diameter $BC$, we have $\angle ICB = \angle IFB$, because both are inscribed in the arc $IB$. Thus, $\angle ICH = \angle BFH = \angle HFG = \angle HCG$. Therefore, since the right triangles $CHG$ and $CHJ$ have the same angles and a common leg, they are congruent, resulting in $HJ = GH = 2 \text{ cm}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_82072e76ed187b592b1cg-33.jpg?height=746&width=667&top_left_y=259&top_left_x=782)