In a trapezium 𝐴𝐵𝐶𝐷, diagonals 𝐴𝐶 and 𝐵𝐷 intersect at 𝑂. if 𝐴𝐵 = 3𝐶𝐷, then find the ratio of areas of triangles 𝐶𝑂𝐷 and 𝐴𝑂𝐵.

                                                   (OR)

In the given figure, 𝑃𝐴 and 𝑃𝐵 are tangents from a point 𝑃 to the circle with centre 𝑂. At the point 𝑀, another tangent to the circle is drawn cutting 𝑃𝐴 and 𝑃𝐵 at 𝐾 and N. Prove that the perimeter of △𝑃𝑁𝐾 = 2𝑃𝐵.

It is given that,in a trapezium 𝐴𝐵𝐶𝐷, diagonals 𝐴𝐶 and 𝐵𝐷 intersect at point 𝑂. 𝐴𝐵 = 3𝐶𝐷 ⇢ (ⅰ) 

In △𝐴𝑂𝐵 and △𝐶𝑂𝐷 

⇒∠𝐴𝑂𝐵 = ∠𝐶𝑂𝐷 [since𝐴𝐵 ∥ 𝐶𝐷, ∠𝐴𝑂𝐵 and ∠𝐶𝑂𝐷 are vertically opposite angles] 

⇒∠𝐴𝐵𝑂 = ∠𝐶𝐷𝑂 [ 𝐴𝐵 ∥ 𝐶𝐷 & 𝐵𝐷 is traversal, alternate angles are equal] 

Therefore, by 𝐴𝐴 similarity, △𝐴𝑂𝐵 and △𝐶𝑂𝐷 are similar We know that the ratio of two similar triangles is equal to the square of the ratio of their corresponding sides.

$\therefore \left[\frac{\text{ area of }\mathrm{△}COD}{\text{ area of }\mathrm{△}AOB}\right]={\left[\frac{CD}{AB}\right]}^2$

$\begin{array}{r}={\left[\frac{CD}{3CD}\right]}^2\text{ (substituting from (i)) }\\ ={\left[\frac{1}{3}\right]}^2\\ =\frac{1}{9}\end{array}$

Hence, the ratio of area of triangles 𝐴𝑂𝐵 and 𝐶𝑂𝐷 is 1: 9.

                                                      (OR)

It is given that 𝑃𝐴 and 𝑃𝐵 are tangents from a point 𝑃 to the circle with centre O and at the point M, another tangent to the circle is drawn cutting 𝑃𝐴 and 𝑃𝐵 at 𝐾 and 𝑁 

⇒ 𝐾𝑁 = 𝐾𝑀 + 𝑀𝑁 [tangents drawn from an external point to a circle are equal]

⇒ 𝐾𝑁 = 𝐾𝐴 + 𝐵𝑁 [since 𝐾𝑀 = 𝐾𝐴 𝑎𝑛𝑑 𝑀𝑁 = 𝐵𝑁] 

Perimeter of △𝑃𝑁𝐾 = 𝑃𝑁 + 𝐾𝑁 + 𝑃𝐾 

= 𝑃𝑁 + 𝐵𝑁 + 𝐾𝐴 + 𝑃𝐾

 = 𝑃𝐵 + 𝑃𝐴 = 2𝑃𝐵 [since 𝑃𝐴 = 𝑃𝐵] 

Perimeter of △𝑃𝑁𝐾 = 2PB