Chapter 2.1, Problem 7PSA

To determine

To Prove: $\angle C=\angle D$

Explanation of Solution

Given information: $\angle ACB={90}^{\circ },AD\perp BD.$

Formula Used: Angle in a semi circle is ${90}^{\circ }$ and its converse

Proof:

  $\begin{array}{l}\angle ACB={90}^{\circ }\mathrm{....}(given)\\ AD\perp BD\mathrm{....}(given)\\ \therefore \angle ADB={90}^{\circ }\end{array}$

  $\therefore$ AB is a diameter of a circle as it makes angle ${90}^{\circ }$ at the centre .

  $\therefore \angle ACB=\angle ADB$ ( Each is equal to ${90}^{\circ }$ )