In Triangle ABC, internal bisector of ∠ABC, external bisector of ∠ACB meet at P.

[Ganesh Ujwal]

In Triangle ABC, internal bisector of ∠ABC, external bisector of ∠ACB meet at P.

In Triangle ABC, the internal bisector of [FONT=MathJax_Main]∠[FONT=MathJax_Math-italic]A[/FONT][FONT=MathJax_Math-italic]B[/FONT][FONT=MathJax_Math-italic]C[/FONT][/FONT] and the external bisector of [FONT=MathJax_Main]∠[FONT=MathJax_Math-italic]A[/FONT][FONT=MathJax_Math-italic]C[/FONT][FONT=MathJax_Math-italic]B[/FONT][/FONT] meet at P. If [FONT=MathJax_Main]∠[FONT=MathJax_Math-italic]B[/FONT][FONT=MathJax_Math-italic]A[/FONT][FONT=MathJax_Math-italic]C[/FONT][/FONT] = 40 what is the measure of [FONT=MathJax_Main]∠[FONT=MathJax_Math-italic]B[/FONT][FONT=MathJax_Math-italic]P[/FONT][FONT=MathJax_Math-italic]C[/FONT][/FONT]?

1) 40

2) 20

3) 70

4) 35

5) None of these.

 

[Dr.Peterson]

In Triangle ABC, the internal bisector of [FONT=MathJax_Main]∠[FONT=MathJax_Math-italic]A[/FONT][FONT=MathJax_Math-italic]B[/FONT][FONT=MathJax_Math-italic]C[/FONT][/FONT] and the external bisector of [FONT=MathJax_Main]∠[FONT=MathJax_Math-italic]A[/FONT][FONT=MathJax_Math-italic]C[/FONT][FONT=MathJax_Math-italic]B[/FONT][/FONT] meet at P. If [FONT=MathJax_Main]∠[FONT=MathJax_Math-italic]B[/FONT][FONT=MathJax_Math-italic]A[/FONT][FONT=MathJax_Math-italic]C[/FONT][/FONT] = 40 what is the measure of [FONT=MathJax_Main]∠[FONT=MathJax_Math-italic]B[/FONT][FONT=MathJax_Math-italic]P[/FONT][FONT=MathJax_Math-italic]C[/FONT][/FONT]?

1) 40

2) 20

3) 70

4) 35

5) None of these.

We ask you to show what you have tried, so that we can tell what help you need. Do you at least have any ideas?

One thing you might do is to label angle ACB as x, and write expressions for other angles. Eventually, you should be able to write an equation for the sum of angles in triangle PBC, from which you can find the answer. There may be a much simpler method, perhaps using some theorem you have recently learned, but this at least works!

 

[Ganesh Ujwal]

I think answer is 40. I draw the diagram like this:

 

[Deleted member 4993]

I think answer is 40. I draw the diagram like this:
View attachment 9322

View attachment 9322 In your drawing, you showed:

angle APC = 40° = angle ABC

That cannot happen - think about it!!

 

[Ganesh Ujwal]

That's all I can do. Any clue?

If I really know the answer, I won't post the question at first place.

May be my question is wrong.

 

[Dr.Peterson]

I think answer is 40. I draw the diagram like this:
View attachment 9322

You have misread the problem! Do you know what an external bisector is?

In Triangle ABC, the internal bisector of [FONT=MathJax_Main]∠[FONT=MathJax_Math-italic]A[/FONT][FONT=MathJax_Math-italic]B[/FONT][FONT=MathJax_Math-italic]C[/FONT][/FONT] and the external bisector of [FONT=MathJax_Main]∠[FONT=MathJax_Math-italic]A[/FONT][FONT=MathJax_Math-italic]C[/FONT][FONT=MathJax_Math-italic]B[/FONT][/FONT] meet at P. If [FONT=MathJax_Main]∠[FONT=MathJax_Math-italic]B[/FONT][FONT=MathJax_Math-italic]A[/FONT][FONT=MathJax_Math-italic]C[/FONT][/FONT] = 40 what is the measure of [FONT=MathJax_Main]∠[FONT=MathJax_Math-italic]B[/FONT][FONT=MathJax_Math-italic]P[/FONT][FONT=MathJax_Math-italic]C[/FONT][/FONT]?

 

[Dr.Peterson]

I think answer is 40. I draw the diagram like this:
View attachment 9322

Next time you write, be sure to explain your thinking as well as state an answer.

Note also that, in addition to drawing two internal angle bisectors, you labeled the vertices in a way that doesn't match the problem.

For a definition of external (or exterior) angle bisectors, see here.