In Fig. 4.145, If

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 equation iii)

Then, ΔCED ~ΔABC.