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