8. In the fig.4.142 given, if AB ⊥ BC, DC ⊥ BC, and DE ⊥ AC, prove that ΔCED ∼ ΔABC

Solution:

Given:

AB ⊥ BC,

DC ⊥ BC,

DE ⊥ AC

To prove:

 ΔCED∼ΔABC

From ΔABC and ΔCED

∠B = ∠E = 90^o   [given]

∠BAC = ∠ECD     (alternate angles since, AB || CD with BC as transversal)

Therefore, ΔCED∼ΔABC (AA similarity)