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