In the diagram below, CBFD is a circle such that BC||FD - NSC Technical Mathematics - Question 8 - 2022 - Paper 2

Here, we know:

1. $\angle C_A = \angle F_i = 37^\circ$ (using the tangent-chord theorem).
2. Therefore, $\angle C_2 = \angle F_i = 37^\circ$ (alternate segment theorem with regards to the circle).

So, the size of $\hat{C_2}$ is $37^\circ$.

Since $\angle D_A = 37^\circ$ (angles in the same segment), we have:

1. $\angle F_i = 37^\circ$ (already proven in the previous step).
2. Also, $MD = MF$ (sides opposite equal angles in triangle).

Thus, it follows that $MD = MF$.

For CHDM to be cyclic:

1. $\angle H_i = 74^\circ$ (exterior angle)
2. $\angle D_i = 37^\circ$ (same as angle previously calculated).
3. Since $\angle C_A + \angle M = 180^\circ$ (sum of opposite angles in a cyclic quadrilateral), it proves that CHDM is indeed a cyclic quadrilateral.