# trisecting diagonal

• Feb 13th 2013, 01:06 AM
geniusgarvil
trisecting diagonal
ABCD is a cyclic quadrilateral such that AB is 7 , BC is 24 , CD is 15 and DA is 20 units respectively. points E and F lie on AC such that AE=EF=FC . what is the area of the quadrilateral BEDF.
• Feb 13th 2013, 01:46 AM
earboth
Re: trisecting diagonal
Quote:

Originally Posted by geniusgarvil
ABCD is a cyclic quadrilateral such that AB is 7 , BC is 24 , CD is 15 and DA is 20 units respectively. points E and F lie on AC such that AE=EF=FC . what is the area of the quadrilateral BEDF.

1. By first inspection:

$AD^2+CD^2 = AB^2+BC^2$

That means AC = 25 and ABCD consists of 2 right triangles with a common hypotenuse.

2. The long way: In a cyclic quadrilateral the interior angles of opposite vertices add up to 180° (or pi).

Use Cosine rule:

$AC^2 = AD^2 + DC^2 - 2 \cdot AD \cdot DC \cdot \cos(\angle(ADC))$

$AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(-\angle(ADC))$

Solve this system of equations for x and AC.

3. Let a denote the area. Then

$a_{BEDF} = \frac13 \cdot a_{ABCD}$