图中显示了一个等腰三角形 ABC，其中角 ABC = p°，且 BA = BC。 点 X 使得角 BCX = p°/2。 D 是 AB 延长线与 CX 延长线的交点。 约翰声称 AD 将是以 B 为圆心的圆的直径。 判断约翰的说法是否正确。

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 $B$.