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Question
In the given figure, DB⊥BC, DE⊥AB and AC⊥BC.
Prove that BEDE=ACBC.
In the given figure, DB⊥BC, DE⊥AB and AC⊥BC.
Prove that BEDE=ACBC.
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Solution
In △BED and △ACB, we have:
∠BED = ∠ACB = 90°
∵ ∠B + ∠C = 180°
∴ BD ∥ AC
∠EBD = ∠CAB (Alternate angles)
Therefore, by AA similarity theorem, we get:
△BED~△ACB
⇒ BE/AC = DE/BC
⇒ BE/DE = AC/BC
This completes the proof. -
∠BED = ∠ACB = 90°
∵ ∠B + ∠C = 180°
∴ BD ∥ AC
∠EBD = ∠CAB (Alternate angles)
Therefore, by AA similarity theorem, we get:
△BED~△ACB
⇒ BE/AC = DE/BC
⇒ BE/DE = AC/BC
This completes the proof.
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