Question

D and E are the points on the sides AB and AC respectively of a $△$ ABC such that: AD = 8 cm, DB = 12 cm, AE = 6 cm and CE = 9 cm. Prove that BC = 5/2 DE.

Answer 1

If a line intersect the two sides in distinct points and the two sides are divided in the same ratio, then the line drawn is parallel to the third side of the triangle.

$\frac{AE}{EC}=\frac{6}{9}=\frac{2}{3}$.......(1)

$\frac{AD}{DB}=\frac{8}{12}=\frac{2}{3}$......(2)

As $\left(1\right)=\left(2\right)$

So $ED\parallel CB$

As $ED\parallel CB$

$\angle AED=\angle ACB$

$\angle ADE=\angle ABC$

So by $AAA$ property,

$△AED\sim △ACB$

Hence $\frac{AE}{AC}=\frac{ED}{CB}$

$\frac{AE}{AE+EC}=\frac{ED}{CB}$ $[\therefore AC=AE+EC]$

$\frac{6}{6+9}=\frac{ED}{CB}$

$\frac{6}{15}=\frac{2}{5}=\frac{ED}{CB}$

$\therefore CB=\frac{5}{2}ED$ $\left[henceproved\right]$