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. |