$\Delta ABC$ is an isosceles triangle in which $AB = AC$. Side $BA$ is produced to $D$ such that $AD = AB$ (see figure). Show that $\angle BCD$ is a right angle.

(N/A) In $\Delta ABC$,since $AB = AC$,we have $\angle ABC = \angle ACB$ (angles opposite to equal sides). Let $\angle ABC = \angle ACB = x$. Then $\angle BAC = 180^\circ - 2x$.
In $\Delta ACD$,since $AD = AB$ and $AB = AC$,we have $AD = AC$. Thus,$\angle ADC = \angle ACD$ (angles opposite to equal sides). Let $\angle ADC = \angle ACD = y$.
Since $BA$ is produced to $D$,$BD$ is a straight line. Therefore,$\angle BAC + \angle CAD = 180^\circ$.
Since $\angle CAD = 180^\circ - 2y$ (from $\Delta ACD$),we have $(180^\circ - 2x) + (180^\circ - 2y) = 180^\circ$.
$360^\circ - 2(x + y) = 180^\circ$.
$2(x + y) = 180^\circ$.
$x + y = 90^\circ$.
Since $\angle BCD = \angle ACB + \angle ACD = x + y$,we conclude $\angle BCD = 90^\circ$.