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Relationship between Unequal Sides of Triangle and the Angles Opposite to It.

In a Δ ABC,...

Question

In a $Δ A B C, I f A C > A B$ and the bisector of $∠ A$ meets $B C$ at $E$ , then

A

C E > B E

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B

C E < B E

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C

C D > B E

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D

C D < B E

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Solution

The correct option is A $C E > B E$
$G i v e n △ A B C, A C > A B . T h e b i s e c t o r o f A m e e t s B C a t E . C o n s t r u c t i o n : T a k e a p o i n t F o n A C, s u c h t h a t A F = A B . J o i n B F a n d E F . L e t B F i n t e r s e c t A E a t O . △ A B F i s a n i s o s c e l e s t r i a n g l e (S i n c e A B = A F, b y c o n s t r u c t i o n) A E b i s e c t s ∠ A (g i v e n) ∴ A E i s t h e p e r p e n d i c u l a r b i s e c t o r o f B F, s o a n y p o i n t o n A E i s e q u i d i s t a n t f r o m B a n d F . I n p a r t i c u l a r, B E = F E S o, i n △ B E F, ∠ E B F = ∠ E F B = α, (a n g l e s o p p o s i t e e q u a l s i d e s) S o, ∠ B E O = 90 = ∠ F E O (L e t a n g l e = γ) T h u s A E a l s o b i s e c t s ∠ B E F . N o w, b e c a u s e i n a t r i a n g l e, e x t e r i o r a n g l e i s t h e s u m o f i n t e r i o r o p p o s i t e a n g l e s, s o ∠ E F C = θ + γ a n d e x ∠ B = θ + γ S o, ∠ E F C = e x t ∠ B . B u t ∠ B = 2 θ + ∠ C, s o ∠ E F C = 2 θ + ∠ C . ∴ ∠ E F C > ∠ C S o, C E > F E \Rightarrow C E > B E (S i n c e B E + F E)$

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