Hello, Finroe!

As TPHacker pointed out, the two triangles are __not__ congruent . . .

In triangle ABC, BC is produced to E and CB is produced to D.

AC = AB and the ratio of sides is: $\displaystyle \frac{AB}{DB} \,= \,\frac{CE}{AB}$

Prove triangle ABD is similar to ACE. Code:

A
*
*/ \ *
* / \ *
* / \ *
* / \ *
* / \ *
* / \ *
* / \ *
* - - - * - - - - - - - * - - - - - - - - - - - *
D B C E

From $\displaystyle \frac{AB}{DB} \,= \,\frac{CE}{AB}$ and $\displaystyle AB = AC$, we have: .$\displaystyle \frac{AB}{DB} \,= \,\frac{CE}{AC}$

Since $\displaystyle \angle ABC$ and $\displaystyle \angle ACB$ are base angles of an isosceles triangle: .$\displaystyle \angle ABC\,=\,\angle ACB$

. . Hence: .$\displaystyle \angle ABD \,= \,\angle ACE$ . (Supplements of equal angles are equal.)

Therefore, by "sas": .$\displaystyle \Delta ABD \sim \Delta ACE$