Proof : We are given two triangles ABC and PQR such that ∆ABC ~ ∆ PQR (see Fig.).
Proof that : ${ar(ABC)}/{ar(PQR)}=({AB}/{PQ})^2= ({BC}/{QR})^2 = ({CA}/{RP})^2$

Proof
For finding the areas of the two triangles, we draw altitudes AM and PN of the triangles.
Now, ar (ABC) = $1/2{BC × AM}$
and ar (PQR)= $1/2{QR × PN}$
So, ${ar (ABC)}/{ar (PQR)}={1/2{BC × AM}}/{1/2{QR × PN}}$

${ar (ABC)}/{ar (PQR)}={BC × AM}/{QR × PN}.......................(1)$

Now, in and ,


∠ B = ∠ Q (As ∆ ABC ~ ∆ PQR)

and ∠ M = ∠ N / (Each is of 90°)

So, ∆ ABM ~ ∆ PQN (AA similarity criterion)

Therefore, ${AM}/{PN}={AB}/{PQ} ............................(2)$

Also, ∆ ABC ~ ∆ PQR (Given)

So, ${AB}/{PQ}={BC}/{QR}={CA}/{RP} ....................(3)$

Therefore, ${ar (ABC)}/{ar (PQR)}={AB}/{PQ}{AM}/{PN} ........[From (1) and (3)] $

${ar (ABC)}/{ar (PQR)}={AB}/{PQ}.{AB}/{PQ}..........from (2)$

${ar (ABC)}/{ar (PQR)}=({AB}/{PQ})^2$

Now using (3), we get
${ar (ABC)}/{ar (PQR)}=({AB}/{PQ})^2=({BC}/{QR})^2=({CA}/{RP})^2$

Example: In Fig., the line segment XY is parallel to side AC of ∆ ABC and it divides the triangle into two parts of equal areas. Find the ratio ${AX}/{AB}$

Solution:
We have XY || AC ...............(Given)
So, ∠ BXY = ∠A and ∠ BYX = ∠C (Corresponding angles)
Therefore,∆ABC ~ ∆ XBY ........(AA similarity criterion)
So, ${ar(ABC)}/{ar(XBY)}=({AB}/{XB})^2$

Also, ar (ABC) = 2 ar (XBY)............. (Given)

${ar(ABC)}/{ar(XBY)}=2/1$

Therefore, from (1) and (2),

${({AB}/{XB})^2=2/1}$

${({AB}/{XB})=√2/1}$

${({XB}/{AB})=1/√2}$

${(1-{XB}/{AB})=1-1/√2}$

${({AB-XB}/{AB})=(√2-1)/√2}$

${({AX}/{AB})=(√2-1)/√2=(2-√2)/2}$