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