Question

In the adjoining figure, points $A,B,C,D$ lie on a circle. $AD=24$ and $BC=12$. What is the ratio of the area of $\triangle CBE$ to that of $\triangle ADE$? 

 

• A. $1:4$
• B. $1:2$
• C. $1:3$
• D. Insufficient data

Hint:

When two chords intersect inside a circle, triangles formed by corresponding chord-pairs are often similar; area ratios then follow from the square of the chord-length ratio.

Answer 1

The Correct Option is A

Let $E$ be the intersection of chords $BA$ and $CD$ (as in the figure). Claim: $\triangle ADE \sim \triangle CBE$.
Angles between intersecting chords inside a circle are equal when they intercept the same pair of arcs: \[ \angle AED = \angle CEB,\qquad \angle ADE = \angle CBE. \] Hence the triangles are similar with the correspondence \[ \triangle ADE \sim \triangle CBE\quad\Rightarrow\quad \frac{AD}{CB}=\frac{\text{scale of sides}}{}. \] Therefore, the ratio of their areas equals the square of the side ratio: \[ \frac{[CBE]}{[ADE]}=\left(\frac{CB}{AD}\right)^{\!2} =\left(\frac{12}{24}\right)^{\!2}=\frac{1}{4}. \]  Final Answer: \(\boxed{1:4}\)