I am stumped on the following question:

In triangle ABC , AD=DB , DE is parallel to BC. The area of Triangle ABC is 40. What is the area of triangle ADE

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I know Thales theorem must be applied here . But I cant figure how to ? Any suggestions ?


Since $DE$ is parallel to $BC$, hence $\triangle ABC$ is similar to $\triangle ADE$. Now since, $AD=DB \implies AB=2AD\implies $ altitude from $A$ to $DE$ is half of altitude from $A$ to $BC$ and $BC=2DE$. Given, Area$(\triangle ABC)=40\implies \frac{1}{2}(\perp from A to BC)(BC)=40\implies \frac{1}{2}(2*\perp from A to DE)(2*DE)=40\implies \frac{1}{2}(\perp from A to DE)(DE)=10$.

Hence Area$(\triangle ADE)=10$.

  • $\begingroup$ Thanks that was straight-forward. $\endgroup$ – Rajeshwar Jul 26 '12 at 12:05

I'm not sure which theorem of Thales you are thinking of, but since you know that $\overline{DE}$ is parallel to $\overline{BC}$, $\overline{AB}$ is a transversal, so we have that $\angle ADE=\angle ABC$. Obviously, $\angle DAE=\angle BAC$, therefore, $\triangle ABC$ is similar to $\triangle ADE$. Since $\overline{AD}$ is half the length of $\overline{AB}$, the area of $\triangle ADE$ is one quarter of the area of $\triangle ABC.$


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