Question

From a point A outside a circle, a secant ABC is drawn cutting the circle at B and C, and a tangent AD touching it at D. A chord DE is drawn equal in length to chord DB. Prove that triangles ABD and CDE are similar.

Answer 1

Given that $AD$ is a tangent and $ED=BD$

According to the alternate segment theorem, angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

$⟹\angle ADB=\angle DCB-\left(1\right)$

Since $ED=BD$

$\angle ECO=\angle BCD–\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$\angle ADB=\angle ECD–\left(3\right)$

Also, $BCDE$ is a cyclic quadrilateral so, opposite angles are supplementary

$\angle DBC+\angle DEC=180$

$\angle DEC=180-\angle DBC–\left(4\right)$

$\angle DBC+\angle DBA=180$

$\angle DBA=180-\angle DBC$

From $\left(4\right)$ and $\left(5\right)$

$\angle DEC=\angle DBA–\left(6\right)$

In triangles $\Delta DECand\Delta DBA$

$\angle ADB=\angle ECD,\angle DEC=\angle DBA$

By AA similarity $△ABD$ and $△CDE$ are similar