SOLUTION: in triangle ABC,the bisector of interior angle at vertex B and the bisector of exterior angle at vertex A intersect each other at point P.Prove that:2 angle APB=angle C.

Algebra ->  Parallelograms -> SOLUTION: in triangle ABC,the bisector of interior angle at vertex B and the bisector of exterior angle at vertex A intersect each other at point P.Prove that:2 angle APB=angle C.      Log On


   

Question 490241: in triangle ABC,the bisector of interior angle at vertex B and the bisector of exterior angle at vertex A intersect each other at point P.Prove that:2 angle APB=angle C.
Answer by Edwin McCravy(20064) About Me  (Show Source):
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in triangle ABC,the bisector of interior angle at vertex B and the bisector of exterior angle at vertex A intersect each other at point P.Prove that:2 angle APB=angle C.




To make things a little easier I labeled the measures of the two equal 
halves of the interior angle at B with little d's and the measures of 
the two equal halves of the exterior angle at A with little f's.

1. ∠CAQ = ∠C + ∠ABC     because in triangle ABC, the exterior angle at A is
                        equal to the sum of the remote interior angles.

2. 2f = ∠C + 2d         using the small letters to replace the angle 
                        measures in the preceding step.

3. ∠PAQ = ∠P + ∠PBA     because in triangle ABP, the exterior angle at A is
                        equal to the sum of the remote interior angles.

4. f = ∠P + d           using the small letters to replace the angle
                        measures in the preceding step.

5. 2(∠P + d) = ∠C + 2d  Substituting the right side of 4 for f in 2 

6. 2∠P + 2d = ∠C + 2d   Distributive principle from preceding step.

7. 2∠P = ∠C             Subtracting 2d from both sides of preceding equation.

Edwin