Task B-3.6. A given isosceles triangle has a base length of $6 \mathrm{~cm}$ and legs of length 5 cm. A circle whose diameter is one of the legs of the triangle intersects the other two sides of the triangle at points $E$ and $F$. Determine the length of the segment $|E F|$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_bde1d6b20cdcb90302c4g-16.jpg?height=838&width=1202&top_left_y=481&top_left_x=475)

Sketch.

$|A C|=|B C|=5 \text{ cm}, |A B|=6 \text{ cm}$

$\measuredangle C E B=90^{\circ}$ (angle subtended by the diameter of the circle)

(1 point)

We conclude that point E is the foot of the altitude of the given isosceles triangle.

Since triangle $\triangle A B C$ is isosceles, $\measuredangle C A B=\measuredangle C B A=\alpha$.

Quadrilateral $E B C F$ is cyclic, hence

$$
\measuredangle E B C + \measuredangle C F E = 180^{\circ}
$$

Then,

$$
\measuredangle A F E = 180^{\circ} - \measuredangle C F E = \measuredangle E B C = \alpha
$$

We conclude that triangle $\triangle A E F$ is isosceles, so $|E F| = |A E| = 3 \text{ cm}$.