In figure, $E$ is any point on median $A D$ of a $△ A B C$ . Show that $a r (A B E) = a r (A C E)$

Open in App

Solution

Median of a triangle divides it into two triangles of equal area.

In $△ A B C, A D$ is the median, so $A r (△ A D B) = A r (△ A D C)$ ..........(i)

In $△ B E C, E D$ is the median, so $A r (△ B E D) = A r (△ C E D)$ ............(ii)

$(i) - (i i)$
$⟹ A r (△ A D B) - A r (△ B E D) = A r (△ A D C) - A r (△ C E D)$

$∴ A r (△ A B E) = A r (△ A C E)$

Suggest Corrections

Similar questions

Δ A B C

E

A D

Δ

A B C

(A B E) = ar (A C E)

In the given figure, E is any point on median AD of a ΔABC. Show that

ar (ABE) = ar (ACE)

E

A D

△ A B C

a r e a (A B E) = a r e a (A C E)

In the given figure, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar (ACE) [2 MARKS]

Join BYJU'S Learning Program