[Thales' Theorem and the Proportional Segments Theorem] [Orthogonal (Rectangular) Projection $]$

In rectangle $A B C D$, $A B=3$, $B D=6$. On the extension of the bisector $B L$ of triangle $A B D$, a point $N$ is taken, and point $L$ divides segment $B N$ in the ratio $10: 3$, counting from point $B$. Which is greater: $B N$ or $C L$?
Since $BL$ is the bisector, then $AL:LD=1:2$.

Let $M$ be the projection of point $N$ onto $AD$. Then $AM:DM=\frac{13}{10}:\left(3-\frac{13}{10}\right)=13:17$, which means $AM<DM$.

Therefore, $BN<CL$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-17.jpg?height=363&width=464&top_left_y=2563&top_left_x=797)

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-18.jpg?height=172&width=1449&top_left_y=220&top_left_x=13)

The bisector of the angle adjacent to angle $C$ of triangle $ABC$ intersects the extension of side $AB$ beyond point $B$ at point $D$, and the bisector of the angle adjacent to angle $A$ intersects the extension of $BC$ beyond point $C$ at point $E$. It is known that $DC=CA=AE$. Find the angles of triangle $ABC$.

## Hint

Let $\angle BAC=\alpha$ and set up an equation in terms of $\alpha$.

## Solution

Let $\angle BAC=\angle ADC=\alpha$. The external angle of the isosceles triangle $ACD$ at vertex $C$ is $2\alpha$. Therefore, the external angle of triangle $ABC$ at vertex $C$ is $4\alpha$. Thus, $\angle CAE=180^\circ-8\alpha$. The external angle of triangle $ABC$ at vertex $A$ is twice the angle $CAE$, so $\angle BAC=180^\circ-2(180^\circ-8\alpha)=16\alpha-180^\circ$. From the equation $16\alpha-180^\circ=\alpha$, we find that $\alpha=12^\circ$. Then $\angle ACB=180^\circ-4\alpha=132^\circ$,

$\angle ABC=180^\circ-(12^\circ+132^\circ)=36^\circ$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-18.jpg?height=566&width=1217&top_left_y=1100&top_left_x=429)

## Answer

$12^\circ, 36^\circ, 132^\circ$.