## Task 3 - 340733

If $A B C$ is a triangle that is not obtuse, let $D$ be the foot of the perpendicular from $C$ to $A B$; $E$ be the image of $D$ under reflection in $A C$, and $F$ be the image of $D$ under reflection in $B C$.

a) What is the area and perimeter of the pentagon $A B F C E$, assuming $A B=7 \mathrm{~cm}$ and $C D=4 \mathrm{~cm}$?

b) What is the angle $\angle A C B$, if - different from in a) - it is assumed that there is a line on which the three points $E, C$, and $F$ lie?

Also prove that from this assumption it follows that the quadrilateral $A B F E$ is a trapezoid!
(a) Since in reflection every point of the line being reflected remains fixed and every triangle is transformed into a congruent triangle, $\triangle A C E = \triangle A C D$ and $\triangle B C F = \triangle B C D$ (see figure a).

Therefore, the area of $A B F C E$ is twice as large as that of $A B C$. This area is given as $12 A B \cdot C D = 14 \text{ cm}^2$, so $A B F C E$ has an area of $28 \text{ cm}^2$.

Furthermore, it follows that $A E = A D$ and $B F = B D$, thus $A E + B F = A B$, and $C E = C F = C D$. Therefore, the perimeter of $A B F C E$ is $2 \cdot A B + 2 C D = 22 \text{ cm}$.

(b) With the notations $\angle B A C = \alpha$ and $\angle A B C = \beta$, the interior angle theorem for the triangles $A D C$ and $B D C$, and due to the reflections, we have $\angle A C E = \angle A C D = 90^\circ - \alpha$ and $\angle B C F = \angle B C D = 90^\circ - \beta$.

For $\angle A C B = \gamma$, given the condition $\angle E C F = 180^\circ$ (see figure b), we have $90^\circ - \alpha + \gamma + 90^\circ - \beta = 180^\circ$.

Since $180^\circ - \alpha - \beta = \gamma$ (interior angle theorem for triangle $ABC$), it follows that $2 \gamma = 180^\circ$, thus $\gamma = 90^\circ$. Furthermore, because $\triangle A C E = \triangle A C D$ and $\triangle B C F = \triangle B C D$, we have $\angle A E C = \angle B F C = 90^\circ$, making the segments $A E$ and $B F$ both perpendicular to $E F$ and thus parallel to each other. Therefore, $A B F E$ is proven to be a trapezoid.