[ Inscribed Quadrilaterals (Miscellaneous) ]

Quadrilateral $ABCD$ is inscribed in a circle. Point $X$ lies on its side $AD$, such that $BX \parallel CD$ and $CX \parallel BA$. Find $BC$, if $AX = 3/2$ and $DX = 6$.

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Triangles $A B C, B X C$ and $X C D$ are similar.

## Solution

Let the angles at vertices $A, B$, and $X$ of triangle $A B X$ be denoted as $\alpha, \beta$, and $\gamma$ respectively. Since $C X \| B A$ and $B X \| C D$, then

$\angle D C X = \angle B X C = \angle A B X = \beta, \quad \angle C D X = \angle B X A = \gamma$ and $\angle C X D = \angle B A X = \alpha$.

$\angle C B A = 180^{\circ} - \angle C D A = 180^{\circ} - \gamma = \alpha + \beta$, and since $\angle A B X = \beta$, then $\angle C B X = \alpha$. Therefore, triangles $A B C, B X C$ and $X C D$ are similar. Multiplying the equalities $B C: A X = C X: B X$ and $B C: D X = B X: C X$ term by term, we get $B C^2 = A X \cdot D X = 9$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8c5c78c6462ff090b828g-22.jpg?height=686&width=704&top_left_y=0&top_left_x=677)

## Answer

3.

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