9.5. Square $A B C D$ and isosceles right triangle $A E F$ $\left(\angle A E F=90^{\circ}\right)$ are positioned such that point $E$ lies on segment $B C$ (see figure). Find the angle $D C F$.

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The square $A B C D$ and the isosceles right triangle $A E F$ with $\angle A E F = 90^{\circ}$ are arranged so that point $E$ lies on the segment $B C$. Determine the measure of angle $D C F$.
Answer: $45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c23c631c658e2131dd58g-2.jpg?height=334&width=443&top_left_y=558&top_left_x=1526)

Solution. First method. Let $P$ be the foot of the perpendicular dropped from point $F$ to line $B C$ (see Fig. 9.5a). Since $\angle F E P=90^{\circ}-\angle B E A=\angle E A B$, the right triangles $F E P$ and $E A B$ are congruent (by hypotenuse and acute angle). Therefore, $P F=B E$. Moreover, $B E=B C-C E=$ $=A B-C E=E P-C E=P C$. Thus, $P F=P C$, which means triangle $C P F$ is a right isosceles triangle. Hence, $\angle F C P=45^{\circ}$, and therefore $\angle D C F=45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c23c631c658e2131dd58g-2.jpg?height=329&width=426&top_left_y=1229&top_left_x=381)

Fig. 9.5a

![](https://cdn.mathpix.com/cropped/2024_05_06_c23c631c658e2131dd58g-2.jpg?height=411&width=437&top_left_y=1142&top_left_x=1255)

Fig. 9.5b

Second method. Draw the diagonal $A C$. Since $\angle E C A=\angle E F A=45^{\circ}$, the quadrilateral $E F C A$ is cyclic (see Fig. 9.5b). Then $\angle A C F=\angle A E F=90^{\circ}$. Therefore, $\angle D C F=\angle A C F-$ $-\angle A C D=90^{\circ}-45^{\circ}=45^{\circ}$.

## Grading Criteria:

+ A complete and well-reasoned solution is provided

$\pm$ A generally correct reasoning is provided, with minor inaccuracies or gaps

- Only the answer is provided or the answer is obtained from a specific example
- The problem is not solved or is solved incorrectly