Question

If for acute angled $\Delta ABC$ and $\Delta PQR$ are similar by $ABC\leftrightarrow PQR$ correspondance, then prove that $\frac{A\left(ABC\right)}{A\left(PQR\right)}=\frac{A{B}^2}{P{Q}^2}=\frac{B{C}^2}{Q{R}^2}=\frac{A{C}^2}{P{R}^2}$

Answer 1

We are given two triangles ABC and PQR such that $\Delta ABC\leftrightarrow \Delta PQR$.
Draw altitudes AM and PN of the triangles.
Now, $A\left(ABC\right)=\frac{1}{2}\times BC\times AM$
and $A\left(PQR\right)=\frac{1}{2}\times QR\times PN$
So, $\frac{A\left(ABC\right)}{A\left(PQR\right)}=\frac{\frac{1}{2}\times BC\times AM}{\frac{1}{2}\times QR\times PN}=\frac{BC\times AM}{QR\times PN}$ ............. (1)
In $\Delta ABM$ and $\Delta PQN$,
$\angle B=\angle Q$ (As $\Delta ABC\leftrightarrow \Delta PQR$)
and $\angle M=\angle N$ (Each of ${90}^o$)
So, $\Delta ABM\leftrightarrow \Delta PQN$ (AA similarity criterion)
$\therefore \frac{AM}{PN}=\frac{AB}{PQ}$ .......... (2)
Also $\Delta ABC\leftrightarrow \Delta PQR$
So, $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}$ ......... (3)
$\therefore \frac{ABC}{PQR}=\frac{AB}{PQ}\times \frac{AM}{PN}$ [From (1) and (3)]
$=\frac{AB}{PQ}\times \frac{AB}{PQ}$ [From (2)]
$={\left(\frac{AB}{PQ}\right)}^2$
Now using (3), we get
$\frac{A\left(ABC\right)}{A\left(PQR\right)}={\left(\frac{AB}{PQ}\right)}^2={\left(\frac{BC}{QR}\right)}^2={\left(\frac{CA}{RP}\right)}^2$