8.3. In triangle $A B C$ with $m(<A)=90^{\circ}$, we construct the bisector $C D$ of angle $A C B, D \in(A B)$ and $B E \perp C D, E \in C D$. Knowing that $C D=2 \cdot D E$, find the measures of the acute angles of triangle $A B C$.
Solution:

![](https://cdn.mathpix.com/cropped/2024_06_06_8e09b761ed0e58f60e66g-2.jpg?height=688&width=696&top_left_y=327&top_left_x=66)

Let $D E=x, C D=2 x, A D=m, B D=n$. From the Angle Bisector Theorem $\Rightarrow \frac{B C}{n}=\frac{A C}{m}:=y \Rightarrow B C=n y, A C=m y$. $\triangle B E D \sim \triangle C A D$ (AA) $\Rightarrow \frac{n}{2 x}=\frac{x}{m} \Leftrightarrow m n=2 x^{2}$ and $m(\angle A C D)=m(\angle B C E)=m(\angle E B D):=\alpha$.

$\triangle B E C \sim \triangle D A C$ (AA) $\Rightarrow \frac{n y}{2 x}=\frac{3 x}{m y} \Leftrightarrow m n y^{2}=6 x^{2}$. From (2) and (3) $\Rightarrow y^{2}=3$.

From $\triangle A B C$ by applying the Pythagorean Theorem

we get $(m+n)^{2}+m^{2} y^{2}=n^{2} y^{2} \Leftrightarrow(m+n)^{2}=\left(n^{2}-m^{2}\right) y^{2} \Leftrightarrow(m+n)=(n-m) y^{2} \Leftrightarrow$ $\Leftrightarrow(m+n)=3(n-m) \Leftrightarrow 4 m=2 n \Leftrightarrow n=2 m$. From (2) $\Rightarrow m=x$.

From $\triangle D A C \Rightarrow A D=\frac{1}{2} D C \Rightarrow m(\angle D C A)=30^{\mathrm{O}} \Rightarrow m(\angle C)=60^{\mathrm{O}}, m(\angle B)=30^{\mathrm{O}}$.