[Theorem on the lengths of a tangent and a secant; the product of the entire secant and its external part [Two tangents drawn from one point

A circle passing through point $D$ and touching the sides $A B$ and $B C$ of an isosceles trapezoid $A B C D$ intersects the sides $A D$ and $C D$ at points $M$ and $N$, respectively. It is known that $A M: D M=1: 3, C N: D N=4: 3$. Find the base $B C$, if $A B=7$ and $A D=6$.

#
Apply the tangent-secant theorem.

## Solution

Let $P$ and $Q$ be the points of tangency of the given circle with the sides $AB$ and $BC$ of trapezoid $ABCD$. By the tangent-secant theorem,

$$
A P^{2}=A D \cdot A M=6 \cdot \frac{3}{2}=9
$$

Therefore, $A P=3$. Then,

$$
B Q=B P=A B-A P=7-3=4
$$

By the tangent-secant theorem,

$$
Q C^{2}=C D \cdot C N=7 \cdot 4=28
$$

Therefore, $Q C=2 \sqrt{7}$. Hence,

$$
B C=B Q+Q C=4+2 \sqrt{7}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-08.jpg?height=346&width=466&top_left_y=936&top_left_x=796)

## Answer

$4+2 \sqrt{7}$