Question

In the following figure, if AB ⊥ BC, DC ⊥ BC and DE ⊥ AC, prove that ΔCED ~ ΔABC.

Answer 1

Given: AB ⊥ BC, DC ⊥ BC and DE ⊥ AC

To prove: ΔCED ~ ΔABC

Proof:

∠BAC + ∠BCA = 90°                  …(i) [By angle sum property]

And, ∠BCA + ∠ECD = 90°             …(ii) [DC ⊥ BC given]

Compare equation (i) and (ii)

∠BAC = ∠ECD                           …(iii)

In ΔCED and ΔABC

∠CED = ∠ABC                        [Each 90°]

∠ECD = ∠BAC                         [From (iii)]

Then, ΔCED ~ ΔABC               [By AA similarity]