Question

In the figure, seg $AB$ and seg $AD$ are tangent segments drawn to a circle with centre $C$ from exterior point $A$, then prove that:
$\begin{array}{cc}\angle A=\frac{1}{2}& \left[m\right(\mathrm{arc}BBP)\\ -m\left(\mathrm{arc}BXD\right)]& \end{array}$

Answer 1

Explanation:
Given, a circle with centre $C$,
$AB$ and $AD$ are the tangent on the circle
Therefore, $\angle ABC=\angle ADC={90}^{\circ }$ (radius perpendicular on
tangent)
Step 1:
From quadrilateral $ABCD$, and we know that sum of angles of a quadrilateral is ${360}^{\circ }$.
$\begin{array}{cc}& \angle A+\angle B+\angle C+\angle D=360\\ & \Rightarrow \angle A+90+\angle C+90=360\\ & \Rightarrow \angle A+\angle C=360-180=180\\ & \Rightarrow \angle A+m\left(\mathrm{arc}BXD\right)=180\\ & \Rightarrow \angle A=180-m\left(\mathrm{arc}BXD\right)\end{array}$
Step 2:
Now we consider that,
$$m\left(\mathrm{arc}BXD\right)+m\left(\mathrm{arc}BYD\right)={360}^{\circ }\dots \dots$$
And if we multiply the equation(i) with 2 we get,
$$2\angle A=360-2m\left(\mathrm{arcBXD}\right)\dots \dots \text{(iii)}$$
Now put the value of ${360}^{\circ }$ in equation (iii)we get,
$\begin{array}{cc}& \Rightarrow 2\angle A=m\left(\mathrm{arc}BXD\right)+m\left(\mathrm{arc}BYD\right)-2m\left(\mathrm{arc}BXD\right)\\ & \Rightarrow 2\angle A=m\left(\mathrm{arc}BYD\right)-m\left(BXD\right)\\ & \Rightarrow \angle A=\frac{1}{2}\left[m\right(\mathrm{arcBY}D)-m(\mathrm{arc}BXD\left)\right]\end{array}$
Final answer:
Hence, we proved that $\angle A=\frac{1}{2}\left[m\right(\mathrm{arcBY}D)-m(\mathrm{arc}BXD\left)\right]$