#### Question

In Figure below, if AB ⊥ BC, DC ⊥ BC and DE ⊥ AC, Prove that Δ CED ~ ABC.

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]

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#### APPEARS IN

Solution In Figure Below, If Ab ⊥ Bc, Dc ⊥ Bc and De ⊥ Ac, Prove that δ Ced ~ Abc. Concept: Triangles Examples and Solutions.