$A$ is one of the points where the two circles intersect.
$AB$ is a chord in the left circle and it's tangent in point A to the right circle.
$AC$ is a chord in the right circle and it's tangent in point A to the left circle.
$BC$ intersects the two circles in point $D$ and $E$.
Prove $AD = AE$
I tried to use the theorem that says that a tangent line to a circle is perpendicular to the radius connected to the tangent line in its point (they don't tell where is the center of the circle).
I tried to use the theorem that says that the angle between the tangent line and a chord is equal to the inscribed angle subtending in the same chord. Nor don't know how to use it.
Any help?
let $\angle BAD = \alpha,\ \angle DAE = \beta, \ \angle EAC = \gamma.$ we will use the fact that the angle between a chord and the tangent at the point of touch is the angle made by the chord.
that is, we have $$\angle DBA = \beta + \gamma, \angle ECA = \alpha + \beta. $$ now, $$\angle EDA = \angle DBA + \angle BAD = \alpha+\beta+\gamma. $$ in the same way $$\angle BEA = \angle ECA + \angle EAC = \alpha+\beta+\gamma. $$
therefore the triangle $ADE$ is isosceles and $AD = AE. $