# Divide a triangle into 2 parts equal area by a line parallel to its base.

• Mar 24th 2013, 07:32 PM
Jammix
Divide a triangle into 2 parts equal area by a line parallel to its base.
A proof or brief explanation why would be nice too! Thank you!!
• Mar 25th 2013, 06:35 AM
Soroban
Re: Divide a triangle into 2 parts equal area by a line parallel to its base.
Hello, Jammix!

Quote:

Divide a triangle in two equal areas by the line parallel to its base.
Code:

    -      *     :      *: *     :    * :h  *     H    *  :    *     :  * * * * * * *     :  *      b      *     - * * * * * * * * * *               B
The large triangle has base $B$ and height $H.$
The small triangle has base $b$ and height $h.$

We want: . $\tfrac{1}{2}bh \:=\:\tfrac{1}{2}\left(\tfrac{1}{2}BH\right) \quad\Rightarrow\quad bh \:=\:\tfrac{1}{2}BH\;\;[1]$

From similar triangles: . $\frac{b}{h} \,=\,\frac{B}{H} \quad\Rightarrow\quad b \,=\,\tfrac{B}{H}h$

Substitute into [1]: . $\left(\tfrac{B}{H}h\right)h \:=\:\tfrac{1}{2}BH \quad\Rightarrow\quad h^2 \:=\:\tfrac{1}{2}H^2$

Therefore: . $h \:=\:\frac{H}{\sqrt{2}}$