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

Since $\angle C_A = 37^{\circ}$ and $\angle F_2$ is in the same segment as $\angle C_A$, we can conclude:

$\angle C_2 = 37^{\circ}$ (as they subtend the same arc BC|FD).

From 8.2, we know angles $\angle C_1$ and $\angle F_1$ are equal. Hence, triangles M.D.F and M.D.C are similar:

$\angle F_1 = \angle C_1 = 37^{\circ}$ Thereby showing that side MD is opposite to angle $\angle D$ and side MF is opposite to angle $\angle C$:

By demonstrating that the angles $\angle H_M + \angle D_M = 180^{\circ}$, we can conclude that quadrilateral CHDM is cyclic. Since:

• $\angle H_M$ and $\angle D_M$ are exterior angles of triangle CHD.
• Therefore, the cyclic nature is evident:

$CHDM is a cyclic quad.$