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Question: ${\triangle}ABC\ \text{is an isosceles triangle, given}\ \overline{AD}=2\overline{EC}\ \text{, compute the value of}\cos{B}$

My Attempt: $\cos{B} = \frac{\overline{BD}}{\overline{AB}}=\frac{\overline{BE}}{\overline{BC}}$

That's all I can do...

I noticed that ${\triangle}ADB\sim{\triangle}CEB$, but I don't know how to use it

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2 Answers 2

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By the similarity $$\frac{AB}{CB}=\frac{AD}{CE}=2$$ $AB = 2 BC $ then $\cos B =1/4$.

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Another approach (other than similarity) is to calculate the area of $\triangle ABC$ in two different ways, i.e. $\frac 12 EC \cdot AB = \frac 12 AD \cdot BC$, allowing you to deduce that $AB = 2BC = 4BD$ and $\cos \angle ABC = \frac {BD}{AB} = \frac 14$.

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