Try this beautiful Problem on Geometry from Trapezium from (AMC 10 A, 2009).

**Trapezium **– AMC-10A, 2009- Problem 23

Convex quadrilateral $A B C D$ has $A B=9$ and $C D=12$. Diagonals $A C$ and $B D$ intersect at $E, A C=14$, and $\triangle A E D$ and $\triangle B E C$ have equal areas. What is $A E ?$

,

- $11$
- $12$
- $13$
- $14$
- $6$

**Key Concepts**

Geometry

**quadrilateral**

Similarity

## Suggested Book | Source | Answer

#### Suggested Reading

Pre College Mathematics

#### Source of the problem

AMC-10A, 2009 Problem-23

#### Check the answer here, but try the problem first

$6$

## Try with Hints

#### First Hint

Given that Convex quadrilateral $A B C D$ has $A B=9$ and $C D=12$. Diagonals $A C$ and $B D$ intersect at $E, A C=14$, and $\triangle A E D$ and $\triangle B E C$ have equal areas. we have to find out the length of \(AE\).

Now if we can show that \(\triangle AEB\) and \(\triangle DEC\) are similar then we can find out \(AE\)?

Can you find out?

#### Second Hint

Given that area of \(\triangle AED\) and area of \(\triangle BEC\) are equal. Now area of \(\triangle ABD\) = area of \(\triangle AED\) + \(\triangle ABE\)

Area of \(\triangle ABC\) = area of \(\triangle AEB\) + \(\triangle BEC\)

Therefore area of \(\triangle ABD\)= area of \(\triangle ABC\) [as area of \(\triangle AED\) and area of \(\triangle BEC\) are equal]

Since triangles $A B D$ and $A B C$ share a base, they also have the same height and thus $\overline{A B} || \overline{C D}$ and $\triangle A E B \sim \triangle C E D$ with a ratio of 3: 4

Can you finish the problem?

#### Third Hint

Therefore $A E=\frac{3}{7} \times A C,$ so $A E=\frac{3}{7} \times 14=6$

## Other useful links

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=y-8Ru_qKDxk