Theorem of Sines [ Ratios of linear elements of similar triangles ]

In an acute triangle $A B C$, heights $A P$ and $C Q$ are drawn from vertices $A$ and $C$. Find the side $A C$, given that the perimeter of triangle $A B C$ is 15, the perimeter of triangle $B P Q$ is 9, and the radius of the circumscribed circle of triangle $B P Q$ is $9 / 5$.
Triangle $P B Q$ is similar to triangle $A B C$ with a similarity coefficient of $\cos \angle B$, and since the ratio of the perimeters of similar triangles is equal to the similarity coefficient, then $\cos \angle B=3 / 5$. Let $R$ be the radius of the circumscribed circle of triangle $A B C$, then $R=5 / 3 \cdot 9 / 5=3$. Therefore, $A C=2 R \sin \angle B=2 \cdot 3 \cdot 4 / 5=24 / 5$.

|  | $\left[\begin{array}{l}\text { Auxiliary circle }\end{array}\right]$ | Difficulty: 4 |
| ---: | :--- | ---: |
| Problem $\underline{52503}$ Topics: | Inscribed angle subtending a diameter $]$ | Grades: 8,9 |
| $[\underline{\text { Inscribed quadrilaterals (other) }}]$ |  |  |

Author: Mikhaylovsky V.

The diagonals of a convex quadrilateral are perpendicular to each other. Prove that the four projections of the intersection point of the diagonals onto the sides of the quadrilateral lie on one circle.

## Hint

The intersection point of the diagonals, its projections on adjacent sides, and the common point of these sides lie on one circle.

## Solution

Let $O$ be the intersection point of the diagonals $A C$ and $B D$ of the quadrilateral $A B C D ; M, N, P, Q$ be the projections of point $O$ on $A B, B C, C D$ and $A D$. Then

$$
\angle Q M O+\angle Q P O=\angle Q A O+\angle Q D O=90^{\circ}
$$

(since points $M$ and $Q$ lie on a circle with diameter $A O$, and points $P$ and $Q$ lie on a circle with diameter $O D$).

Similarly, $\angle O M N+\angle O P N=90^{\circ}$. Therefore,

$$
\angle Q P N+\angle Q M N=180^{\circ}
$$

Thus, quadrilateral $M N P Q$ is cyclic.

![](https://cdn.mathpix.com/cropped/2024_05_06_c317f3f1f50d1d30304bg-43.jpg?height=477&width=594&top_left_y=1322&top_left_x=732)

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