Q16 of 56 Page 6

In Fig., if AB⊥BC, DC⊥BC and DE⊥AC, prove that Δ CED∼Δ ABC.

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)


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