If D is a point on the side AB of ∆ABC such that AD : DB = 3.2 and E is a point on BC such that DE || AC.
If D is a point on the side AB of ∆ABC such that AD : DB = 3.2 and E is a point on BC such that DE || AC. Find the ratio of areas of ∆ABC and ∆BDE.
Given: In ΔABC, D is a point on side AB such that AD : DB= 3 : 2. E is a point on side BC such that DE || AC.
To find: $\frac{\Delta \mathrm{ABC}}{\Delta \mathrm{BDE}}$
In ΔABC,
$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{3}{2}$
Since DE\|AC,
$\frac{E C}{E B}=\frac{3}{2}$
According to converse of basic proportionality theorem if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.
Hence DE || AC
In ΔBDE and ΔABC,
∠BDE=∠A Corresponding angles
∠DBE=∠ABC Common
So, ∆BDE~∆ABC (AA Similarity)
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
Let AD = 2x and BD = 3x.
Hence
$\frac{A r(\triangle \mathrm{ABC})}{A r(\triangle \mathrm{BDE})}=\frac{\mathrm{AB}^{2}}{\mathrm{BD}^{2}}$
$=\frac{(\mathrm{BD}+\mathrm{DA})^{2}}{(\mathrm{BD})^{2}}$
$=\frac{(3 x+2 x)^{2}}{(2 x)^{2}}$
$=\frac{(5 x)^{2}}{(2 x)^{2}}$
$\frac{A r(\Delta \mathrm{ABC})}{\operatorname{Ar}(\Delta \mathrm{BDE})}=\frac{25}{4}$