In the rhombus $ABCD$, points $E$ and $F$ lie on the extension of side $AB$ beyond $B$. The tangents drawn from $E$ and $F$ to the incircle of the rhombus intersect the line $AD$ at points $E'$ and $F'$. Determine the ratio $DE': DF'$, given that $BE: BF = \lambda: \mu$.
Solution. First, we prove a lemma from which the solution to our problem follows easily.

Lemma. If the tangent from point $P$ on the extension of side $AB$ of rhombus $ABCD$ beyond $B$ to the inscribed circle of the rhombus intersects line $AD$ at point $P'$, and the center of the inscribed circle of the rhombus is $O$, then

$$
P^{\prime} D: D O=O B: B P
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_d7f506deeab45c8f428fg-1.jpg?height=368&width=464&top_left_y=434&top_left_x=817)

Let the angles of triangle $APP'$ be denoted as $2\varphi, 2\alpha, 2\beta$ as shown in the figure. Since the inscribed circle of the rhombus is also the inscribed circle of triangle $APP'$, $O$ is the intersection of the angle bisectors of the triangle. Therefore, $\angle D P^{\prime} O = \angle O P^{\prime} P = \beta$ and $\angle B P O = \angle O P P^{\prime} = \alpha$. The third angle of triangle $OPP'$ is thus

$$
\angle P O P^{\prime} = 180^{\circ} - (\alpha + \beta) = 90^{\circ} + \varphi
$$

Since $ABCD$ is a rhombus, triangle $BCD$ is isosceles with the angle at $C$ being $2\varphi$. Therefore, $\angle C B D = \angle C D B = \frac{180^{\circ} - 2\varphi}{2} = 90^{\circ} - \varphi$. The opposite sides of the rhombus are parallel, so $\angle P^{\prime} D C = \angle C B P = \angle D A B = 2\varphi$, which gives us

$$
\angle P^{\prime} D O = \angle P^{\prime} D C + \angle C D B = 2\varphi + (90^{\circ} - \varphi) = 90^{\circ} + \varphi = \angle P^{\prime} O P
$$

and

$$
\angle O B P = \angle D B C + \angle C B P = (90^{\circ} - \varphi) + 2\varphi = 90^{\circ} + \varphi = \angle P^{\prime} O P
$$

This means that triangles $P^{\prime} D O$ and $O B P$ are similar to triangle $P^{\prime} O P$ because two corresponding angles are equal. Therefore, the two triangles are also similar to each other, and the ratios of their corresponding sides are equal, which implies the desired statement $P^{\prime} D: D O = O B: B P$.

Returning to our original problem: Applying the statement of the lemma to points $E$ and $F$, we get that $E^{\prime} D: D O = O B: B E$ and $F^{\prime} D: D O = O B: B F$. Dividing these two equalities by each other, we obtain the desired ratio: $E^{\prime} D: F^{\prime} D = B F: B E = \mu: \lambda$.

Using the work of Nóra Katalin Sándor (Pápai Ref. Koll. Gimn., 12th grade)