In an isosceles trapezoid $A B C D$ the bases $A D=12, B C=6$, the height is 4. The diagonal $A C$ divides the angle $B A D$ of the trapezoid into two parts. Which one is larger?
Consider triangle $A B C$.

## Solution

Let $P$ and $Q$ be the projections of vertices $B$ and $C$ onto the base $A D$ of trapezoid $A B C D$. Then $A P=D Q=1 / 2(A D-B C)=3$. By the Pythagorean theorem, $A B^{2}=A P^{2}+B P^{2}=25$.

In triangle $A B C$, side $B C$ is greater than side $A B$ (6>5). Therefore, $\angle B A C>\angle B C A=\angle C A D$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c271881b12104a3b803eg-28.jpg?height=374&width=698&top_left_y=2357&top_left_x=680)

## Answer

Angle $BAC$.

## [ Right triangles (miscellaneous). $\quad]$  Problem 53902 Topics: [ Sum of angles in a triangle. External angle theorem.]  [ Properties and criteria of isosceles triangles. ]

On the extensions of the hypotenuse $A B$ of right triangle $A B C$ beyond points $A$ and $B$, respectively, points $K$ and $M$ are taken such that $A K=A C$ and $B M=B C$. Find the angle $M C K$.

## Hint

$\angle C A B$ and $\angle C B A$ are external angles of isosceles triangles $C A K$ and $C B M$.

## Solution

$\angle C A B$ and $\angle C B A$ are external angles of isosceles triangles $C A K$ and $C B M$. Let $\angle C A B=2 \alpha, \angle C B A=2 \beta$. Then $\angle C K M=\alpha$ and $\angle C M K=\beta$. Therefore, $\angle M C K=180^{\circ}-(\alpha+\beta)=180^{\circ}-45^{\circ}=135^{\circ}$.

## Answer

$135^{\circ}$.