Question

.triangle ABC is inscribed in circle with centre O. seg AD is a diameter. tangent to the circle at point D intersects seg AB in X and AC in Y.
to prove: AB x AX = AC x AY.

Answer 1

Triangle ABC is inscribed in circle with centre O. seg AD is a diameter. tangent to the circle at point D intersects seg AB in X and AC in Y. To prove: AB x AX = AC x AY.

**Answer:**

Let $\angle CDY$ be $x{}^{\circ}$, then:

1) $\angle DAC = x{}^{\circ}$;

Similarity, $\angle CBD = x{}^{\circ}$. (i)

$\angle ABD = 90{}^{\circ}$ (ii) Angle in the semicircle.

2) $\angle ABC = (90 - x){}^{\circ}$, From (i) and (ii). Also, in $\Delta DCY$, $\Delta DCY = 90{}^{\circ}$, $\Delta ACD = 90{}^{\circ}$

And $\angle DCY = x{}^{\circ}$.

3) $\angle CYD = (90 - x){}^{\circ}$. Angle sum property;

4) $\Delta ABC$ and $\Delta AYX$: $\angle ABC = \angle AYX = (90 - x){}^{\circ}$.

Also $\angle A = \angle A$.

5) $\Delta ABC \sim \Delta AYX$ (AA criterion), then $\frac{AB}{AY} = \frac{AC}{AX} \rightarrow AB \times AX = AC \times AY$.

Hence proved.