Question

In the given figure, CD is a direct common tangent to two circles intersecting each other at A and B, then $\angle CAD+\angle CBD=..........$

• A. ${120}^{\circ }$
• B. ${90}^{\circ }$
• C. ${360}^{\circ }$
• D. ${180}^{\circ }$

Answer 1

The correct option is D ${180}^{\circ }$

Join AB.

Now, $\angle ADC=\angle ABD$ [Angle in the alternate segment]

$\angle ACD=\angle ABC$ [Angle in the alternate segment]

In triangle CAD,

$\angle CAD+\angle ADC+\angle ACD={180}^{\circ }$ [Angle sum property]

$\angle CAD+(\angle ABD+\angle ABC)={180}^{\circ }$

$\angle CAD+\angle CBD={180}^{\circ }$

So, option D is the answer.