Question

$ABC$ is a right triangle with $\angle A={90}^{\circ }$. Let a circle touch tangent $\overline{AB}$ at A and tangent $\overline{BC}$ at some point D. Suppose the circle intersects $\overline{AC}$ again at E and $CE=3cm,CD=6cm$, find the measure of BD

• A. $9cm$
• B. $3\sqrt{5}cm$
• C. $3cm$
• D. $2cm$

Answer 1

The correct option is A $9cm$

Given $\angle BAC={90}^{\circ }$

$AB$ is tangent at $A$

$BDC$ is tangent at $D$

$CE=3,CD=6$

According to the tangent-secant theorem the length of tangent segment squared equals the product of secant segment and its external segment.

$⟹AC\times CE=C{D}^2$

$AC=\frac{6^2}{3}=12$

$AE=12–CE=9=d=2r$

Let $O$ be the center of the circle

$OA=OD=r$

$AB=BD=l$, tangents drawn from $B$

Since $\angle A=90$

$B{C}^2=A{C}^2+A{B}^2$

$⟹(l+CD{)}^2=A{C}^2+l^2$

$⟹l^2+6^2+12l={12}^2+l^2$

$⟹12l=108$

$BD=l=9cm=$ length of tangent