Question

In $\triangle ABC$, seg $DE \parallel BC$. If $2A(\triangle ADE) = A(\triangle DBCE)$, find $AB : AD$ and show that $BC = \sqrt{3 \, DE$.}

Hint:

In similar triangles, the ratio of areas is equal to the square of the ratio of their corresponding sides. Always use this relation to compare medians, sides, or heights.

Answer 1

Step 1: Given information.
In $\triangle ABC$, $DE \parallel BC$. Therefore, $\triangle ADE \sim \triangle ABC$ by Basic Proportionality Theorem (BPT).
Step 2: Ratio of areas of similar triangles.
For two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides. \[ \frac{A(\triangle ADE)}{A(\triangle ABC)} = \left(\frac{AD}{AB}\right)^2 \] Step 3: Given condition.
It is given that $2A(\triangle ADE) = A(\triangle DBCE)$. Now, \[ A(\triangle ABC) = A(\triangle ADE) + A(\triangle DBCE) \] Substitute $A(\triangle DBCE) = 2A(\triangle ADE)$: \[ A(\triangle ABC) = A(\triangle ADE) + 2A(\triangle ADE) = 3A(\triangle ADE) \] Step 4: Substitute in the area ratio.
\[ \frac{A(\triangle ADE)}{A(\triangle ABC)} = \frac{1}{3} \] Hence, \[ \left(\frac{AD}{AB}\right)^2 = \frac{1}{3} \] \[ \frac{AD}{AB} = \frac{1}{\sqrt{3}} \] \[ AB : AD = \sqrt{3} : 1 \] Step 5: Relation between BC and DE.
Since $\triangle ADE \sim \triangle ABC$, \[ \frac{BC}{DE} = \frac{AB}{AD} = \sqrt{3} \] \[ \therefore BC = \sqrt{3} \, DE \] Step 6: Conclusion.
Hence, $AB : AD = \sqrt{3} : 1$ and $BC = \sqrt{3} \, DE$.
Correct Answer: $AB : AD = \sqrt{3} : 1$ and $BC = \sqrt{3} \, DE$