Question

Given above, the point D divides the side BC of ∆ABC in the ratio m:n. Prove that area of ∆ABD : area of ∆ADC = m:n​

Answer 1

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Given:

Point D divides the side BC of ∆ABC in the ratio $m:n$.

To Prove:

Area of ∆ABD : Area of ∆ADC = $m:n$.

Proof:

Area of Triangle = $\frac{1}{2}\times Base \times Height$

Area of ∆ABD = $\frac{1}{2}\times BD \times Height$   ------ 1

Area of ∆ADC = $\frac{1}{2} \times DC \times Height$   ------ 2

Now:

1 : 2 = $\frac{1}{2}\times m \times Height$  :  $\frac{1}{2}\times n \times Height$  = $m$ : $n$

[Height is same for both Triangles]

Hence Proved.

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