Question

Prove that a median divides a $△$ into two ${△}^{\prime }s$ of equal area.

Answer 1

Given $\Delta ABC$

With $AD$ as the median

$\therefore BD=CD=\frac{1}{2}BC$

To prove

area of $\Delta ADB=$ area of $\Delta ADC$

Draw line $AN\perp BC$

To find area, we use

Area of triangle $=\frac{1}{2}$ Base $\times$ Altitude

In $\Delta ABD$

$BD$ is the base & $AN$ is Altitude

area of $\Delta ABD=\frac{1}{2}\times$ Base $\times$ Altitude

$=\frac{1}{2}\times BD\times AN$

In $\Delta ADC$

area of $\Delta ADC=\frac{1}{2}\times$ Base $\times$ Altitude

$=\frac{1}{2}\times DC\times AN$

As $BD=DC$

Area of $\Delta ADC=\frac{1}{2}\times BD\times AC$

$\therefore Ar\left(\Delta ABD\right)=Ar\left(\Delta ADC\right)$