Question

In the following figure, if m(arc DXE) = $100{}^\circ$ and m(arc AYC) = $40{}^\circ$ , find $\angle DBE$ .

Answer 1

Hint:Notice that ACED is a cyclic quadrilateral, i.e., the diagonals meet at the centre of the circle. Also, use the property of the circle that the angle at the centre is twice the corresponding angle at the circumference. Finally, use the property that the exterior angle of a triangle is equal to the sum of the opposite interior angles of a triangle.

Complete step-by-step answer:
Let us start the above question by making some constructions according to our needs. We draw the diagonals of the cyclic quadrilateral ACED, and we also know that the meet point of the diagonals of a cyclic quadrilateral is the same as the centre of the circle passing through all the vertices of the cyclic quadrilateral.

Now we will use the theorem that the angle at the circumference is half of the angle at the centre. Also, it is given in the question that m(arc DXE) = $100{}^\circ$ and m(arc AYC) = $40{}^\circ$ . So, we can say:
$\angle AEB=\dfrac{1}{2}\times m\left( arc\text{ AYC} \right)=\dfrac{1}{2}\times 40{}^\circ =20{}^\circ$
$\angle EAD=\dfrac{1}{2}\times m\left( arc\text{ }DXE \right)=\dfrac{1}{2}\times 100{}^\circ =50{}^\circ$
Now according to the property that sum of exterior angle of a triangle is sum of the opposite interior angles, we get
$\angle DBE+\angle AEB=\angle EAD$
Now if we put the known values from the above results, the equation becomes:
$\angle DBE+20{}^\circ =50{}^\circ$
$\Rightarrow \angle DBE=30{}^\circ$
Therefore, the measure of $\angle DBE$ in the above diagram is equal to $30{}^\circ$ .

Note: Remember that the theorems related to cyclic quadrilateral are not valid for any regular quadrilateral. Also, it is prescribed to learn all the basic theorems related to circles as they are regularly used in the questions related to circles and cyclic quadrilaterals, as we did in the above question.