In ΔABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in X. Arrange AX, BX and CX in descending order.

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Solution

Given: In ΔABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in X.

$In ∆ A B X, ∵ ∠ B A X > ∠ A B X ∴ B X > A X . . . (i)$

$Similarly, in ∆ A C X, ∵ ∠ A C X > ∠ X A C ∴ A X > C X . . . (ii)$

$From (i) and (i i), we get B X > A X > C X$

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