Question

Show that a median of a triangle divides it into two triangles of equal area.

Answer 1

Hey...

Here’s the answer... :)

To prove : area of triangle ABD is equal to the area of triangle ACD...

Construction: AM perpendicular to BC...

Now, BD = CD ( D is the median of the side BC ).... (1)

So.., area of triangle ABD = 1/2 * BD * AM...(2)

And also..., area of triangle ACD = 1/2 * CD * AM...(3)

So.., from the equation (3)... we can also say that....

Area of triangle ACD = 1/2 * BD * AM... (4) [Since BD = CD]

So from (4)..., we can conclude that area of the triangles ABD and ACD are equal...

Hence proved!

Hope it helps...

Pls mark as brainliest.. :)

Answer 2

Given : In ΔABC, AD is the median of the triangle.

To prove : ar ΔABD = ar ΔADC

Construction : Draw AP ⊥ BC.

Proof : ar ΔABC = $\sf \frac{1}{2}$ × BC × AP...... (i)

ar ΔABD = $\sf \frac{1}{2}$ × BD × AP

ar ΔABD = $\sf \frac{1}{2} \times \frac{BC}{2} \times AP$

[AD is the median of ΔABC]

ar ΔABD = $\sf \frac{1}{2}$ × ar ΔABC.. (ii)

ar ΔADC = $\sf \frac{1}{2}$ × DC × AP

ar ΔADC = $\sf \frac{1}{2} \times \frac{BC}{2} \times AP$

ar ΔADC =$\sf \frac{1}{2}$ × ar ΔABC.. (iii)

From equation (i), (ii) and (iii)

ar ΔABD = ar ΔADC = $\sf \frac{1}{2}$ ar ΔABC

Hence, it is proved.