Consider ∆ABC ~ ∆DEF
and height of ∆ABC be h1 and ∆DEF be h2
`"AB"/"DE" = "BC"/"EF" = "AC"/"DF"` ....corresponding sides of similar triangles (i)
∠ABC = ∠DEY … corresponding angles of similar triangles(ii)
Consider ∆ABX and ∆DEY
∠AXB = ∠DYE = 90°
From equation (ii)
∠ABC = ∠DEY
∴by AA test for similarity ∆ABX ~ ∆DEY
`"AB"/"DE" = "BX"/"EY" = "AX"/"DY"`....corresponding sides of similar triangles
But from figure AX = h1 and DY = h2
`"AB"/"DE" = "BX"/"EY" = "h"_1/"h"_2`...(iii)
A(∆ABC) = (1/2)×BC×h1
A(∆DEF) = (1/2)×EF×h2
`(A(∆ABC))/(A(∆DEF)) = (AB)/(DE) xx (AB)/(DE) = (AB^2)/(DE^2)` ...(iv)
Squaring equation (i) and using it in (iv)
`(A(∆ABC))/(A(∆DEF)) = (AB^2)/(DE^2) = (BC^2)/(EF^2) = (AC^2)/(DF^2)`
Hence proved
Therefore, the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.