Question

In the figure, AD is median of triangle ABC. Prove that ar(triangle ABD)=ar(triangleACD)​

Answer 1

The median of a triangle divides it into two triangles of equal areas.

Step-by-step explanation:

let ABC be a triangle and Let AD be one of its medians.

In △ABD and △ADC the vertex is common and these bases BD and DC are equal.

Draw AE⊥BC.

Now

$area (\bigtriangleup ABD)= \frac{1}{2}\times base \times altitude\\\\area (\bigtriangleup ABD)= \frac{1}{2}\timesBD \times AE\\\\area (\bigtriangleup ABD)= \frac{1}{2}\times DC \times AE$

but DC and AE is the base and altitude of △ACD

$area (\bigtriangleup ABD)= \frac{1}{2}\times \text{base DC}\times \text{altitude ACD}$

$= ar (\bigtriangleup ACD)$

$ar (\bigtriangleup ABD)= ar (\bigtriangleup ACD)$

Hence the median of a triangle divides it into two triangles of equal areas.

#Learn more:

Given ar ( triangle ABC ) = 32 cm 2, AD is median of triangle ABC and BE is median of triangle ABD. If BO is median of triangle ABE,then ar ( triangle BOE)

https://brainly.in/question/7074175

Answer 2

$\underline { \blue { Solution :}}$

Let ABC be a triangle and Let AD be one of its medians .

In ∆ABD and ∆ADC the vertex is common and these bases BD and DC are equal.

Draw AE perpendicular to BC.

$Now , \pink {area ( \triangle ABD)}\\ = \frac{1}{2} \times base \: BD \times altitude \: of \: \triangle ADB \\= \frac{1}{2} \times BD \times AE \\= \frac{1}{2} \times DC \times AE \: ( Since , BD = DC ) \\= \frac{1}{2} \times base \: DC\times altitude \: of \: \triangle ACD\\= \pink {Area (\triangle ACD )}$

$Hence , \green {Area (\triangle ABD ) \cong Area (\triangle ACD ) }$

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