In a circle with centre O, AB is the diameter and CD is a chord such that ABCD is a trapezium. If $$\angle$$BAC = $$25^\circ$$, then $$\angle$$CAD is equal to:

Solution

As per the question,

ABCD is a trapezium and $$\angle$$BAC = $$25^\circ$$

As we know from the property of trapezium, $$AB\|CD$$

So, $$\angle$$BAC=$$\angle$$ACD$$=25^\circ$$
We know from the circle theorem,
$$\angle$$ACB=$$90^\circ$$
Now, from the cyclic quadrilateral rule,
$$\Rightarrow \angle$$DAB$$+\angle$$DCB$$=180
$$\Rightarrow \angle$$DAC$$\angle$$CAB$$+90+25$$=180$$
$$\Rightarrow \angle$$DAC$$25+90+25$$=180$$
$$\Rightarrow \angle$$DAC$$=180-140$$
$$\Rightarrow \angle$$DAC$$=40^\circ$$