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Question Number 53031 by behi83417@gmail.com last updated on 16/Jan/19

Commented by behi83417@gmail.com last updated on 16/Jan/19

$A B < C B .$ $∡ A B E = ∡ E B F = ∡ F B C$ $. . . . . . s h o w t h a t : \frac{BE}{BF} > 1 .$
	Answered by ajfour last updated on 16/Jan/19

	$l e t t h e t h r e e e q u a l a n g l e s b e θ .$ $\frac{B E}{\sin A} = \frac{A B}{\sin ∠ A E B} &$ $\frac{B F}{\sin C} = \frac{B C}{\sin ∠ C F B}$ $A B < B C \Rightarrow ∠ C < ∠ A \Rightarrow \sin C < \sin A$ $f u r t h e r$ $∠ A E B + ∠ A + θ = ∠ C F B + ∠ C + θ$ $\Rightarrow ∠ C F B > ∠ A E B$ $f r o m e q s . i n f i r s t a n d s e c o n d l i n e$ $\frac{B E}{B F} = \frac{\sin A}{\sin C} \times \frac{\sin ∠ C F B}{\sin ∠ A E B}$ $= (> 1) \times (> 1) > 1$ $h e n c e B E > B F .$
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