Question

In the given figure, CD is a direct common tangent to two circles intersecting each other at A and B, then \[\angle CAD\]+ \[\angle CBD\] = ________
\[
  {\text{A}}{\text{. 120}}^\circ \\
  {\text{B}}{\text{. 90}}^\circ \\
  {\text{C}}{\text{. 360}}^\circ \\
  {\text{D}}{\text{. 180}}^\circ \\
\]

Answer 1

Hint:In this question we had a common tangent given as CD. So here we can apply the alternate segment theorem ( tangent – chord theorem ) and with the help of this theorem we will get the similar angles and further this problem can be solved with the help of angle sum property of a triangle.

Complete step-by-step answer:
Let us assume that AC and AD are the chords of the circles.
Now if AC and AD are the chords and CD is the common tangent
So, by tangent – chord theorem which states that the angle between the chord and a tangent through one of the end points of the chord is equal to the angle in the alternate segment we will get
\[\angle BCD = \angle CAB\] ( 1 )
\[\angle BDC = \angle BAD\] ( 2 )
Now adding equation ( 1 ) and ( 2 )
\[ \Rightarrow \angle BCD + \angle BDC = \angle CAB + \angle BAD\]
\[\]So, \[\angle BCD + \angle BDC = \angle CAD\] ( 3 )
Now in \[\vartriangle \]BCD by applying the angle sum property of a triangle which states that the sum of all the interior angles of the triangle must be equal to \[180^\circ \].
So, \[\angle BCD + \angle BDC + \angle CBD = 180^\circ \] ( by angle sum property of a triangle )
Now putting the value of \[\angle BCD + \angle BDC\] form equation ( 3 ) in above equation
\[ \Rightarrow \]\[\angle CAD + \angle CBD = 180^\circ \]
Hence D is the correct option.

Note:- Whenever we come up with this type of problem we have to first of all find the similar angles or points with the help of various theorems of circle and triangle ( here used tangent – chord theorem and angle sum property of triangle ) inscribed in the figure. And on further proceeding with equating the angles we will get the desired result. This is the easiest and quickest way to solve such types of problems.