The diagram shows a circle with points A, B, D and E on the circle and a point C outside the circle. A, B, D and E are points on the circle such that AD = CD and angle BCD = 28°. Explain with geometrical reasons, why the length of BC equals to BE.

1. Use the given equal lengths to find the angles in triangle $ACD$.

Since $AD=CD$, triangle $ACD$ is isosceles.

Also, $A,B,C$ are collinear, so
$\angle ACD=\angle BCD=28^\circ.$

Hence
$\angle DAC=28^\circ$
and
$\angle ADC=180^\circ-28^\circ-28^\circ=124^\circ.$

2. Find an angle on the circle using a straight-line angle and the same-segment theorem.

Since $C,D,E$ are collinear,
$\angle ADE=180^\circ-\angle ADC=180^\circ-124^\circ=56^\circ.$

Since $A,B,D,E$ lie on the same circle, $\angle ADE$ and $\angle ABE$ are angles in the same segment standing on chord $AE$.

Therefore
$\angle ABE=56^\circ.$

3. Show that triangle $BCE$ is isosceles, hence conclude $BC=BE$.

Since $A,B,C$ are collinear,
$\angle CBE=180^\circ-\angle ABE=180^\circ-56^\circ=124^\circ.$

Also,
$\angle BCE=\angle BCD=28^\circ$
since $C,D,E$ are collinear.

In triangle $BCE$,
$\angle BEC=180^\circ-124^\circ-28^\circ=28^\circ.$

Thus
$\angle BCE=\angle BEC=28^\circ.$

Therefore the opposite sides are equal:
$BC=BE.$