AB, AC are 2 equal chords in circle PA, PB are tangent segments to the circle. Prove that <APB = <BAC??

Question

the first part was to draw a diagram to illustrate this situation...i drew it but am not sure too... plz answer

B;;;;;;;;;
C ;;;;;;;;;;
;;;;;;;;;; p
;;;;;;;;;;;;
'''''''''''''''''''''
..... A
ABC are the tangent chords...P is the point out side the circle....the shape almost looks like ( O> ) so did u get it

Answer 1

simple
ok let x be angle BAC
then angle ACB would be (180-x)/2, since triangle ABC is an issoloces triangle.
Now we know what is ACB which covers the same sector as the perpendicular tangent lines a A & B crossing at the center, this will be twice that of ACB due to a fundemental circle rule.
Now form a kite with points P,A,the center of the circle and B
The total angle of the interior of the kite is 360
Two of the angles are 90 degrees and one of it is (180-x)
all we have to find is the 4th one and that is APB
so-->
let '?' be angle APB

90 +90 + (180-x) + ? = 360
180 - x + ? = 180
-x + ? = 0
x = ?
that is angle BAC = APB

email at: [e-mail] for a diagram and clearer explnation

Answer 2

Please clarify your question. I can't come up with any way of arranging a circle and points A, B, C, and P on the circle, in the circle, or outside the circle where that would be true.