Question

In ΔABC, seg DE || side BC. If 2A(ΔADE) = A(`square`DBCE), find AB : AD and show that BC = `sqrt(3)` DE.

Answer 1

Given: In △ABC, DE ∥ BC. Also 2A(△ADE) = A(DBCE).

To find: AB : AD and prove BC = √3·DE

Using areas and similarity: DE ∥ BC ⇒ △ADE ∼ △ABC

A(△ADE) : A(△ABC) = AD² : AB²

A(DBCE) = A(△ABC) − A(△ADE)

2A(△ADE) = A(DBCE)

2A(△ADE) = A(△ABC) − A(△ADE)

3A(△ADE) = A(△ABC)

A(△ADE) : A(△ABC) = 1 : 3

AD² : AB² = 1 : 3

AB² : AD² = 3 : 1

AB : AD = √3 : 1

Ratio of parallel sides: In similar triangles, DE : BC = AD : AB

DE : BC = 1 : √3

BC = √3 · DE

Hence,

AB : AD = √3 : 1 and BC = √3·DE.