In Fig., If AB ⊥ BC and DE ⊥ AC, prove that ΔCED ~ ΔABC

1.	In Fig., If AB ⊥ BC and DE ⊥ AC, prove that ΔCED ~ ΔABC
Answer» 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.

Answer»

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.

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