Question

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

Answer 1

if in ∆abc triangle
So, $\angle a=\angle a\left(common\right)$

$\angle ade=\angle abc$
corresponding
$\angle aed=\angle acb$
So this ∆abc~ ∆ade
Now by thales theorm
$\frac{ab}{ad}=\frac{bc}{de}$------$ab=20cm$
$\frac{20}{8}=\frac{bc}{de}$

$\frac{5}{2}=\frac{bc}{de}$
$bc=\frac{5}{2}\times de$