Two tangents PA and PB are drawn from an external point P to a circle with center O, such that ∠𝐴𝑃𝐵 = ∠𝑥 and ∠𝐴𝑂𝐵 = 𝑦. Prove that opposite angles are supplementary.

                                               (OR)

In Δ𝐴𝐵C the mid points of of sides 𝐴𝐵, 𝐵𝐶, and CA respectively. Find the ratio of $\mathrm{ar}\left(\mathrm{△}DEF\right)\text{ to }\mathrm{ar}\left(\mathrm{△}ABC\right)$.

It is given that Two tangents 𝑃𝐴 and 𝑃𝐵 are drawn from an external point 𝑃 to a circle with centre O, such that ∠𝐴𝑃𝐵and ∠𝐴𝑂𝐵 = 𝑦 

We know that the tangent at a point to a circle is perpendicular to radius through that point. Therefore 𝑂𝐴⟂𝐴𝑃 and 𝑂𝐵⟂𝐵𝑃 

So ∠𝑂𝐴𝑃 = 90° 

⇒ ∠𝑂𝐵𝑃 = 90° 

⇒ ∠𝑂𝐴𝑃 + ∠𝑂𝐵𝑃 = 90° + 90°…….. (𝑖) 

⇒ = 180°

In a quadrilateral 𝐴𝑂𝐵𝑃 the sum of all angles = 360° 

So ∠𝑂𝐴𝑃 + ∠𝑂𝐵𝑃 + ∠𝐴𝑃𝐵∠ + ∠𝐴𝑂𝐵 = 360° 

⇒ 180° + ∠𝐴𝑃𝐵 + ∠𝐴𝑂𝐵 = 360° [𝑓𝑟𝑜𝑚 (𝑖)] 

⇒ ∠𝐴𝑃𝐵 + ∠𝐴𝑂𝐵 = 360° − 180° 

⇒ ∠𝐴𝑃𝐵 + ∠𝐴𝑂𝐵 = 180° 

Hence, the sum of opposite angles are supplementary.

                                                 (OR)

𝐷, 𝐸, and F are respectively the midpoints of sides 𝐴𝐵, 𝐵𝐶, and CA of Δ𝐴𝐵𝐶

We know that, the line segment joining the midpoints of any two sides of a triangle is half the third side and parallel to it.

Therefore $FD=\frac{1}{2}BC,ED=\frac{1}{2}AC,EF=\frac{1}{2}AB$

In △𝐴𝐵𝐶 and △𝐸𝐹D

$\text{ We have }\frac{AB}{EF}=\frac{BC}{FD}=\frac{AC}{ED}=2$

△𝐴𝐵𝐶 and △𝐸𝐹𝐷 are similar by SSS congruence 

Also, We know that If two triangles are similar, then the ratio of the area of both triangles is equal to the square of the ratio of their corresponding sides.

$\frac{\mathrm{ar}\left(\mathrm{△}ABC\right)}{\mathrm{ar}\left(\mathrm{△}DEF\right)}={\left(\frac{AB}{EF}\right)}^2$

$\begin{array}{r}\Rightarrow \frac{\mathrm{ar}\left(\mathrm{△}ABC\right)}{\mathrm{ar}\left(\mathrm{△}DEF\right)}=4\\ \Rightarrow \frac{\mathrm{ar}\left(\mathrm{△}ABC\right)}{\mathrm{ar}\left(\mathrm{△}DEF\right)}=\frac{1}{4}\end{array}$

Hence, the ratio of 𝑎𝑟(△𝐷𝐸𝐹) to 𝑎𝑟(△𝐴𝐵𝐶) is 1:4