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Properties of Angles Formed by Two Parallel Lines and a Transversal

The sides AB ...

Question

The sides AB and AC of a triangle ABC are produced; and the bisectors of the external angles at B and C meet at P Now if AB >
AC, then PC < PB.
If the above statement is true then mention answer as 1, else mention 0 if false

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Solution

$L e t ∠ A B C = x a n d ∠ A C B = y I f A B > A C t h e n ∠ y > ∠ x (a n g l e s i d e t h e o r e m) S o t h e e x t e r n a l a n g l e w i l l b e ∠ 180 - x, ∠ 180 - y f o r x a n d y r e s p . S o t h e a n g l e m a d e b y P w i t h e B a n d C w i l l b e ∠ P B C = ∠ (\frac{180 - x}{2}) ∠ P C B = ∠ (\frac{180 - y}{2}) S i n c e ∠ y > ∠ x, ∠ 180 - y < ∠ 180 - x S o ∠ P C B < ∠ P B C S o P B < P C$

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and bisector of angle

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If the above statement is true then mention answer as 1, else mention 0 if false

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