The diagram shows an isosceles triangle ABC where angle ABC = p° and BA = BC. Point X is such that angle BCX = p°/2. D is the intersection of AB extended and CX extended. John claims that AD will form a diameter of a circle with centre B. Determine whether John's claim is correct or not.

1. Find the base angle at $C$ in isosceles triangle $ABC$.

Since $BA=BC$, triangle $ABC$ is isosceles with base $AC$.

$\angle BAC=\angle ACB=\frac{180^\circ-p^\circ}{2}=90^\circ-\frac p2.$

2. Show that triangle $BCD$ is isosceles and conclude whether $B$ is the midpoint of $AD$.

Since $D$ lies on $AB$ extended, $BD$ is a straight extension of $BA$.

$\angle CBD=180^\circ-p^\circ.$

Also, since $D$ lies on $CX$ extended and $\angle BCX=\frac p2$, the angle at $C$ in triangle $BCD$ is

$\angle BCD=\frac p2.$

Hence

$\angle BDC=180^\circ-(180^\circ-p^\circ)-\frac p2=\frac p2.$

So $\angle BCD=\angle BDC$, and therefore $BD=BC$.

But $BA=BC$, so $BA=BD$. Thus $B$ is the midpoint of $AD$.

Therefore $AD$ will form a diameter of a circle with centre $B$.