A quadrilateral $$ ABCD $$ is inscribed in a circle such that $$ AB : CD = 2 : 1 $$ and $$ BC : AD = 5 : 4 $$. If $$ AC $$ and $$ BD $$ intersect at the point $$ E $$, then $$ AE : CE $$ equals:

Question

Accepted Answer

![Screenshot 2025-03-30 20.23.09.png](https://res.cloudinary.com/aptidude/image/upload/v1743346398/aptidude_questions/sh281airhhs4yrxdijye.png)

To solve this problem, we use the properties of cyclic quadrilaterals and similar triangles.

#### Step 1: Angle relationships in cyclic quadrilateral
In a cyclic quadrilateral, the angles subtended by the same chord on the circle are equal. Using this property:  
- $$ \angle DAC = \angle DBC $$ (angles subtended by chord DC).  
- $$ \angle ADB = \angle ACB $$ (angles subtended by chord AB).  
- $$ \angle AED = \angle BEC $$ (vertically opposite angles).

Thus, triangles $$ 	riangle AED $$ and $$ 	riangle BEC $$ are similar.

#### Step 2: Ratio from similarity of triangles
From the similarity of triangles:  
$$ \frac{AE}{BE} = \frac{AD}{BC} = \frac{4}{5}, \quad 	ext{and} \quad \frac{ED}{EC} = \frac{CD}{AB} = \frac{1}{2}. $$

Using these ratios:  
$$ \frac{AE}{EC} = \frac{AE}{BE} 	imes \frac{ED}{EC} = \frac{4}{5} 	imes \frac{1}{2} = \frac{8}{5}. $$

#### Final Answer:
The ratio of $$ AE : CE $$ is **8 : 5**.

Answer

Answer

Answer