Krishna
0

Step 1: Make a note of the important points in the question

           GIVEN: ΔABC is an isosceles triangle in which \angle B = 90°

                                   AB = BC


Step 2:  Apply the Pythagoras theorem

               AC^2 = AB^2 + BC^2 [By Pythagoras theorem]

             AC^2= AB^2+ AB^2 [Since AB = BC]

              ∴ AC^2= 2AB^2..................... (1)

               It is also given that ΔABE ~ ΔACD


Step 3: Recall that ratio of areas of similar triangles is equal to ratio of squares of their corresponding sides.

           EXAMPLE:      \frac{Area\ \triangle ABE}{Area\ \triangle ACD}=\frac{AB^2}{AC^2}

                                  \frac{Area\ \ \triangle ABE}{Area\ \ \triangle ACD}=\frac{AB^2}{2AB^2} [from equation (1)]

                                  \frac{Area\ \triangle ABE}{Area\ \triangle ACD}=\frac{1}{2}

                            ∴ ar(ΔABE) : ar(ΔACD) = 1 : 2