# Triangles

• Mar 7th 2011, 09:15 AM
raed
Triangles
Hello,

I need the solution of the following question:

Consider a triangle $\displaystyle \triangle ABC$ and let $\displaystyle X$ be a point on the side $\displaystyle AB$ such that $\displaystyle AX=\frac{1}{3}AB$ and let $\displaystyle Y$ be a point on the side $\displaystyle AC$ such that $\displaystyle CY=\frac{1}{3}AC$. Prove that the area of the triangle $\displaystyle \triangle AXC$ equals the area of the triangle $\displaystyle \triangle BYC$.

Best Regards.
• Mar 7th 2011, 10:20 AM
Soroban
Hello, raed!

Quote:

$\displaystyle \text{Consider }\Delta ABC\text{; let }X\text{ be a point on side }AB\text{ such that: }AX\,=\,\frac{1}{3}AB$
$\displaystyle \text{and let }Y\text{ be a point on the side }AC\text{ such that }CY\,=\,\frac{1}{3}AC$.

$\displaystyle \text{Prove that the area of }\Delta AXC\text{ equals the area of }\Delta BYC.$

Side $\displaystyle \,AB$ is divided in the ratio $\displaystyle 1:2$

Compare $\displaystyle \Delta AXC$ and $\displaystyle \Delta ABC.$

They have the same height $\displaystyle \,h.$

Code:

                    C                     o                   **|*                 * * | *               *  *  |  *             *  *  |h  *           *    *    |    *         *    *    |    *       *      *      |      *   A o * * * o * * * * * * * o B     : - 1 - X - - - 2 - - - :

The base of $\displaystyle \Delta AXC$ is one-third the base of $\displaystyle \Delta ABC.$

Hence: .$\displaystyle (\text{area }\Delta AXC) \;=\;\frac{1}{3}(\text{area }\Delta ABC)$

In a similar fashion, we prove that: .$\displaystyle \text{(Area }\Delta BYC)} \;=\;\frac{1}{3}(\text{area }\Delta ABC})$

And we're done!

• Mar 8th 2011, 10:37 AM
raed
Quote:

Originally Posted by Soroban
Hello, raed!

Side $\displaystyle \,AB$ is divided in the ratio $\displaystyle 1:2$

Compare $\displaystyle \Delta AXC$ and $\displaystyle \Delta ABC.$

They have the same height $\displaystyle \,h.$

Code:

                    C                     o                   **|*                 * * | *               *  *  |  *             *  *  |h  *           *    *    |    *         *    *    |    *       *      *      |      *   A o * * * o * * * * * * * o B     : - 1 - X - - - 2 - - - :

The base of $\displaystyle \Delta AXC$ is one-third the base of $\displaystyle \Delta ABC.$

Hence: .$\displaystyle (\text{area }\Delta AXC) \;=\;\frac{1}{3}(\text{area }\Delta ABC)$

In a similar fashion, we prove that: .$\displaystyle \text{(Area }\Delta BYC)} \;=\;\frac{1}{3}(\text{area }\Delta ABC})$

And we're done!