In a triangle ABC,If a line drawn from vertex A to a point D on BC. If BC = 9 cm and DC = 3 cm, then what is the ratio of the areas of triangle ABD and triangle ADC respectively?

I am thinking about using the property of similar triangles that if two triangles are similar than ratio of their areas is $=$ ratio of square of corresponding sides.


Let $AK$ be an altitude of $\Delta ABC$.

Thus, $$\frac{S_{\Delta ABD}}{S_{\Delta ADC}}=\frac{\frac{1}{2}BD\cdot AK}{\frac{1}{2}DC\cdot AK}=\frac{BD}{DC}=\frac{6}{3}=2$$

  • $\begingroup$ I was thinking of it but couldn't .perceive it $\endgroup$ – Bogorovich Feb 3 '18 at 7:06

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