Q. 55.0( 2 Votes )

# In Δ ABC, the bisector of ∠ A intersects BC in D. The bisector of ∠ADB intersects AB in F and the bisector of ∠ ADC intersects in E. Prove that AF X AB X CE = AE × AC × BF.

Answer :

Given : In Δ ABC, the bisector of ∠ A intersects in D. The bisector of ∠ADB intersects in F and the bisector of ∠ ADC intersects in E

To prove : AF × AB × CE = AE × AC × BF

Proof : in Δ ABC, the bisector of ∠A intersects at D

∴ 1

(∵ in a Δ the bisector of an angle divides the side opposite to the angle in the segments whose lengths are in the ratio of their corresponding sides)

Similarly, In Δ ADB, the bisector of ∠D intersects at F

∴ ………………2

And in Δ ADC , the bisector of ∠D intersects at E

∴ …………..eq(3)

Multiplying 1, 2 and 3, we get

And

(∵ )

∴

∴ AF × BD × CE =AE × CD × BF

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