Question

For two acute angled $△ABC$ and $△PQR$ if $△ABC\sim △PQR$ then prove that $\frac{area\left(△ABC\right)}{area\left(△PQR\right)}=\frac{A{B}^2}{P{Q}^2}=\frac{B{C}^2}{Q{R}^2}=\frac{A{C}^2}{P{R}^2}$.

Answer 1

Construction: Draw $AD$ perpendicular to $BC$ and $PS$ perpendicular to $QR$

As $\Delta ABC\sim \Delta PQR$

So $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}$ (1)

Also $\angle A=\angle P$ , $\angle B=\angle Q$ , $\angle C=\angle R$ (2)

Now consider $\Delta ABD$ and $\Delta PQS$

$\angle B=\angle Q$ (from (2))

$\angle ADB=\angle PSQ=90{}^{\circ }$ (by construction)

So by AA similarity $\Delta ABD\sim \Delta PQS$

Therefore $\frac{AB}{PQ}=\frac{BD}{QS}=\frac{DA}{SP}$ (3)

$area\left(\Delta ABC\right)=\frac{1}{2}\times AD\times BC$ (4)

$area\left(\Delta PQR\right)=\frac{1}{2}\times PS\times QR$ (5)

Dividing (4) by (5)

$\frac{area\left(\Delta ABC\right)}{area\left(\Delta PQR\right)}=\frac{\frac{1}{2}\times AD\times BC}{\frac{1}{2}\times PS\times QR}$ (6)

From (1) and (3)

$\frac{DA}{SP}=\frac{BC}{QR}$ (7)

Substituting (7) in (6)

$\frac{area\left(\Delta ABC\right)}{area\left(\Delta PQR\right)}=\frac{\frac{1}{2}\times BC\times BC}{\frac{1}{2}\times QR\times QR}=\frac{(BC{)}^2}{(QR{)}^2}$

From (1)

$\frac{area\left(\Delta ABC\right)}{area\left(\Delta PQR\right)}=\frac{(BC{)}^2}{(QR{)}^2}=\frac{(AB{)}^2}{(PQ{)}^2}=\frac{(AC{)}^2}{(PR{)}^2}$

Hence proved