In a trapezoid, the bases are equal to $a$ and $b$, the diagonals are perpendicular, and the angle between the lateral sides is $\alpha$. Find the area of the trapezoid.
The sums of the squares of the opposite sides of a quadrilateral with perpendicular diagonals are equal to each other.

## Solution

Let \( AD = b, BC = a \) be the bases of the trapezoid \( ABCD \). Suppose \( b > a \). Draw a line through vertex \( C \) parallel to the lateral side \( AB \) until it intersects the base \( AD \) at point \( F \). Then

\[
DF = b - a, \angle DCF = \alpha.
\]

Denote \( AB = CF = x, CD = y \). Let \( h \) be the height of the trapezoid. Then

\[
h = \frac{2 S_{\triangle DCF}}{DF} = \frac{xy \sin \alpha}{b - a}
\]

By the cosine theorem in triangle \( DCF \), we find that

\[
x^2 + y^2 - 2xy \cos \alpha = (b - a)^2
\]

Since the sums of the squares of the opposite sides of a quadrilateral with perpendicular diagonals are equal to each other, then

\[
x^2 + y^2 = a^2 + b^2
\]

Subtracting the first equation from the second, we get

\[
2xy \cdot \cos \alpha = 2ab \Rightarrow xy = \frac{ab}{\cos \alpha}
\]

Therefore,

\[
\begin{aligned}
& S_{\mathrm{ABCD}} = \frac{1}{2}(BC + AD) h = \frac{a + b}{2} \cdot \frac{xy \sin \alpha}{b - a} = \\
& = \frac{a + b}{2} \cdot \frac{\frac{ab}{\cos \alpha} \cdot \sin \alpha}{b - a} = \frac{ab(a + b) \tan \alpha}{2(b - a)}
\end{aligned}
\]

Similarly for the case \( a > b \).

![](https://cdn.mathpix.com/cropped/2024_05_06_4bea0374a35e8bb7d318g-44.jpg?height=535&width=509&top_left_y=1745&top_left_x=777)

Answer

\(\frac{ab(a + b) \tan \alpha}{2|a - b|}\). Send a comment

On the extension of the chord \( KL \) of a circle with center \( O \), a point \( A \) is taken, and tangents \( AP \) and \( AQ \) (where \( P \) and \( Q \) are the points of tangency) are drawn from it; \( M \) is the midpoint of segment \( PQ \). Prove that \(\angle MKO = \angle MLO\).

## Hint

From the equality \( AO \cdot AM = AK \cdot AL \), it follows that points \( K, L, M \), and \( O \) lie on the same circle.

## Solution

We will prove that points \( K, L, M \), and \( O \) lie on the same circle. This will imply the equality of the specified angles.

Note that points \( A, M \), and \( O \) lie on the same line. Since \( PM \) is the altitude of the right triangle \( APO \) drawn from the vertex \( P \) of the right angle, then \( AP^2 = AO \cdot AM \). On the other hand, by the tangent-secant theorem, \( AP^2 = AK \cdot AL \). Therefore,

\[
AO \cdot AM = AK \cdot AL
\]

Thus, points \( K, L, M \), and \( O \) lie on the same circle.

![](https://cdn.mathpix.com/cropped/2024_05_06_4bea0374a35e8bb7d318g-45.jpg?height=549&width=715&top_left_y=772&top_left_x=680)

| Problem 55753 Topics: | Rotation by \( 90^{\circ} \) |  |  |
| :---: | :---: | :---: | :---: |
|  | Pythagorean Theorem (direct and inverse). |  | Difficulty: 4+ |
|  | Rotation helps solve the problem. |  | Grades: 8,9 |

Author: Kurrindchik Ll.

Point \( P \) is located inside the square \( ABCD \), such that \( AP : BP : CP = 1 : 2 : 3 \). Find the angle \( APB \).

## Hint

Rotate triangle \( APB \) by \( 90^{\circ} \) counterclockwise around vertex \( B \).

## Solution

Assume \( AP = 1, BP = 2, CP = 3 \). Let \( P_1 \) be the image of point \( P \) after a \( 90^{\circ} \) counterclockwise rotation around vertex \( B \). Then \( PBP_1 \) is an isosceles right triangle. Therefore,

\[
\angle BP_1P = 45^{\circ}, P_1P = 2\sqrt{2}
\]

Thus,

\[
P_1P^2 + P_1C^2 = 8 + 1 = 9 = 3^2 = PC^2
\]

This means that triangle \( PP_1C \) is a right triangle, \(\angle PP_1C = 90^{\circ}\). Therefore,

\[
\angle APB = \angle CP_1B = \angle CP_1P + \angle BP_1P = 90^{\circ} + 45^{\circ} = 135^{\circ}.
\]

## Answer

\( 135^{\circ} \).