In Fig., if <span

Given: AB⏊BC

DC ⏊ BC

DE ⏊AC

To prove: ΔCED ~ΔABC

Theorem Used:

If two corresponding angles of two triangles are equal the triangles are said to be similar.

Proof:

In ΔABC,

∠ABC + ∠BAC + ∠BCA = 180° (By angle sum property)

90° + ∠BAC + ∠BCA = 180°

⇒ ∠BAC + ∠BCA = 90°……(i)

As DC ⏊ BC,

∠BCA + ∠ECD = 90° …… (ii)

Compare equation (i) and (ii)

∠BAC = ∠ECD …. (iii)

In ΔCED and ΔABC

∠CED = ∠ABC (Each 90 °)

∠ECD = ∠BAC (From equation iii)

Then, ΔCED ~ΔABC. (By AA similarity)