Question

Answer the question....

Answer 1

REFER TO ATTATCHMENT FOR YOUR ANSWER. ....



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Answer 2

In the given figure,

∠AGB = ∠AFC (each 90°)

⇒ BG ║ CF ( corresponding angles are equal)

Now,

(i) In Δ ABG and Δ DCB

∠AGB = ∠DBC ( each 90°)

∠ABG ∠BCD (corresponding angles since BG & CF are parallel)

∴ Δ ABG is similar to Δ DCB

⇒ BG/BA = BC/BD (BPT) -------(1)

(ii) Now,

In Δ ABG & Δ  ABE

∠ A = ∠ A (common)

∠ABE = ∠AGB (each 90°)

⇒ Δ ABG is similar to Δ ABE

⇒ BG/BA = BE/BA (BPT)

from (1)

⇒ BC/BD = BE/BA (∵ BG/BA = BC/BD = BE/BA)

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