Problem 10.3. Quadrilateral $ABCD$ is such that $\angle ABD = \angle ACD = 90^\circ$, $\angle CAD = 42^\circ$. Rays $CB$ and $DA$ intersect at point $K$. It is known that $BK = 3$, $AD = 6$.

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(a) (2 points) How many degrees does the angle $BKA$ measure?

(b) (2 points) How many degrees does the angle $BAC$ measure?
# Answer:

(a) (2 points) $28^{\circ}$.

(b) (2 points) $34^{\circ}$.

Solution. From the condition, it follows that points $A, B, C, D$ lie on a circle with diameter $A D$. Connect point $B$ to the midpoint $O$ of segment $A D$, which is the center of this circle (Fig. 6). Clearly, $A O=O D=O B=B K=3$.

Triangles $K B O$ and $B O D$ are isosceles. Let $\alpha=\angle B K A=\angle B O K$. The angle $B O K$, equal to $\alpha$, is an exterior angle for triangle $B O D$ and is equal to the sum of the two interior angles at vertices $B$ and $D$, so $\angle O B D=\angle O D B=\frac{\alpha}{2}$.

Since quadrilateral $A B C D$ is inscribed, $\angle C B D=\angle C A D=42^{\circ}$. The angle $C B D$, equal to $42^{\circ}$, is an exterior angle for triangle $K B D$ and is equal to the sum of the two interior angles at vertices $K$ and $D$, so $42^{\circ}=\alpha+\frac{\alpha}{2}$, from which we find $\angle B K A=\alpha=$ $\frac{2}{3} \cdot 42^{\circ}=28^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d4c5d21501884c712855g-5.jpg?height=293&width=616&top_left_y=92&top_left_x=419)

Fig. 6: to the solution of problem 10.3

In the right triangle $A B D$, the sum of the angles at vertices $A$ and $D$ is $90^{\circ}$, so $90^{\circ}=\angle B A C+\angle C A D+\angle A D B=\angle B A C+42^{\circ}+\frac{1}{2} \cdot 28^{\circ}$, from which we find $\angle B A C=34^{\circ}$.