In a triangle ABC, E is the midpoint of median AD, show that $a r e a △ A B E = a r e a △ A C E$

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Solution

AD is the median of $Δ A B C$ .
Therefore, it will divide $Δ A B C$ into two triangles of equal area.
Area $Δ A B D$ = Area $Δ A C D$
$∴ a r e a (Δ A B D) = \frac{1}{2} a r e a (Δ A B C)$ ------------(1)
In $Δ A B D$ , E is the mid-point of AD.
Therefore, BE is the median.
Then Area $Δ B E D$ = Area $Δ A B E$
$∴ a r e a (Δ A B E) = \frac{1}{2} a r e a (Δ A B D)$ ------------(2)
$a r e a (Δ A B E) = \frac{1}{2} \times \frac{1}{2} a r e a (Δ A B C)$
$a r e a (Δ A B E) = \frac{1}{4} a r e a (Δ A B C)$ ...........................(3)
In $Δ A D C$ , E is the mid-point of AD.
Therefore, EC is the median.
Area $Δ A E C$ = Area $Δ C E D$
$∴ a r e a (Δ A C E) = \frac{1}{2} a r e a (Δ A D C)$
$a r e a (Δ A C E) = \frac{1}{2} \times \frac{1}{2} a r e a (Δ A B C)$
$a r e a (Δ A C E) = \frac{1}{4} a r e a (Δ A B C)$ ....................(4)
From (3) and (4) we get
$a r e a (Δ A B E) = a r e a (Δ A C E)$

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