Triangles and Area

Geometry Level pending

In the figure, DE is a parallel line to BC which divides the triangles into two polygons of equal area.

So, find the value of B D A B \dfrac{BD}{AB} .


The answer is 0.293.

Viki Zeta by Viki Zeta , 19, India

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1 solution

Viki Zeta
Sep 9, 2016

D E ∣ ∣ B C Δ A D E ∼ Δ A B C (AA Similarity) ar(ABC) ar(ADE) = ( A B A D ) 2 ar(ABC) 2 ar(ADE) = ( A B A D ) 2 (Since, ar(ADE) = ar(DECB) = 2ar(ABC)) 1 2 = ( A B A D ) 2 1 2 = A B A D 2 = A D A B − 2 = − A D A B 1 − 2 = 1 − A D A B 1 − 2 2 = A B − A D A B D B A B = 2 − 2 2 = 0.293 DE || BC \\ \Delta ADE \sim \Delta ABC \text{ (AA Similarity)} \\ \dfrac{\text{ar(ABC)}}{\text{ar(ADE)}} = (\dfrac{AB}{AD})^2 \\ \dfrac{\text{ar(ABC)}}{2\text{ar(ADE)}} = (\dfrac{AB}{AD})^2 \text{ (Since, ar(ADE) = ar(DECB) = 2ar(ABC))} \\ \dfrac{1}{2} = (\dfrac{AB}{AD})^2 \\ \dfrac{1}{\sqrt[]{2}} = \dfrac{AB}{AD} \\ \sqrt[]{2} = \dfrac{AD}{AB} \\ -\sqrt[]{2} = -\dfrac{AD}{AB} \\ 1 - \sqrt[]{2} = 1 - \dfrac{AD}{AB} \\ 1 - \dfrac{\sqrt[]{2}}{2} = \dfrac{AB - AD}{AB} \\ \dfrac{DB}{AB} = \dfrac{2 - \sqrt[]{2}}{2} = 0.293\\

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