Theorem 6.6 : The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Explanation

Proof `:` The ratio of areas of two triangles is equal to the ratio of the products of their bases and corresponding heights .
`:. (A(Delta ABC ))/(A ( Delta XYZ )) = ( BC xx AD )/( YZ xx XT )`
`:. (A ( Delta ABC))/( A(Delta XYZ)) = ( BC)/( YZ) xx(AD)/(XT ) ` ...(1)
`Delta ABC ~ Delta XYZ `
`(AB)/( XY) = (BC)/( YZ) = ( AC)/( XZ )` ...( Corresponding sides of similar triangles are in proportion ) ...(2)
In `Delta ABD ` and `Delta XYT `,
`/_ ABD ~= /_ XYT ` ...(Given )
`_ADB ~= /_ XTY ` ....(Each of measure `90^(@)`)
`:. DeltaABD ~ Delta XYT ` .....(AA test of similarity )
`:. (AB)/( XY ) = ( AD)/( XT )` ...(Corresponding sides of similar triangles are in proportion ) ...(3)
`:. `from (2) and (3) , we get
`(AB)/( XY) = ( BC)/( YZ) = (AD)/( XT ) `
`:. (BC)/( YZ) = (AD)/( XT ) ` ...(4)
`( A(Delta ABC ))/(A( Delta XYZ)) = ( BC) /( YZ) xx(BC)/( YZ)` ....[From (1) and (4) ]
`:. ( A( Delta ABC ))/(A( Delta XYZ)) = ( BC^(2))/(YZ^(2))` ...(5)
`:. (A( Delta ABC))/(A(Delta XYZ)) = ( AB^(2))/( XY^(2)) = ( BC^(2))/(YZ^(2)) = ( AC^(2))/( XZ^(2)) ` ..[From (1) and( 5) ]`