[Theorem on the lengths of a tangent and a secant; the product of the entire secant and its external segment. Auxiliary similar triangles [Mean proportional in a right triangle

From point $A$, two tangents (with points of tangency $M$ and $N$) and a secant intersecting the circle at points $B$ and $C$, and the chord $MN$ at point $P$, are drawn, with $AB: BC=2: 3$. Find $AP: PC$.

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Let $O$ be the center of the circle, and the line $A O$ intersects the chord $M N$ at point $K$, and $H$ is the projection of point $O$ on the chord $B C$. Then $A B \cdot A C = A M^2 = A O \cdot A K = A H \cdot A P$.

## Solution

Let $O$ be the center of the circle, and the line $A O$ intersects the chord $M N$ at point $K$. Then $A K \perp M N$. From the right triangle $A O M$, we find that

$A M^2 = A O \cdot A K$.

Let $H$ be the projection of point $O$ on the chord $B C$. Then $H$ is the midpoint of $B C$. The right triangles $A K P$ and $A H O$ are similar by two angles, so

$A K : A P = A H : A O$, from which $A O \cdot A K = A H \cdot A P$.

By the theorem of the tangent and the secant, $A M^2 = A B \cdot A C$.

From the obtained equalities, it follows that $A B \cdot A C = A O \cdot A K = A H \cdot A P$.

Let $A B = 2 x, B C = 3 x$. Then $A C = A B + B C = 5 x, A H = \frac{1}{2}(A B + A C) = \frac{7 x}{2}, A P = \frac{A B \cdot A C}{A H} = \frac{20 x}{7}$. Therefore, $P C = A C - A P = 5 x - \frac{20 x}{7} = \frac{15 x}{7}, A P : A C = 4 : 3$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8c5c78c6462ff090b828g-31.jpg?height=494&width=732&top_left_y=1205&top_left_x=663)

## Answer

$4: 3$.