deductive geometry: circle w/ center P circumscribes triangl

[asaver]

I was wondering if anyone could

1)Check my interpretation/drawing of the question

2)Help me solve it

Here is the question:
A circle with centre P circumscribes triangle ABC (ie: touches each vertex). In the triangle: Angle A=75o and Angle B=60o. Construct tangents at points A and B, and extend them to intersect at D-outside the circle.

Show that ABD is an isosceles triangle with a right angle at D.

(Hint: you may need to construct some extra lines within triangle ABC)

Here is what I believe is described in the first paragraph:
http://s9.photobucket.com/albums/a100/alexandersaver/?action=view&current=IMG_0007.jpg

Could anyone help me solve this( as well as where I would put the xtra lines)

PS. I know it can be solved without extra lines but Im not that advanced.
Thanks

 

[TchrWill]

I was wondering if anyone could

1)Check my interpretation/drawing of the question

2)Help me solve it

Here is the question:
A circle with centre P circumscribes triangle ABC (ie: touches each vertex). In the triangle: Angle A=75o and Angle B=60o. Construct tangents at points A and B, and extend them to intersect at D-outside the circle.

Show that ABD is an isosceles triangle with a right angle at D.

1--Let the circle center be O. (If not given the circle center, construct perpendicular bisectors of AC and BC. Their intersection is the circle center O.)
2--Draw BO and AO
3--Angle BCA is equal to 1/2 the intercepted arc BA. With angle BCA being 45º, angle BOA is therefore 90º.
4--The tangents at A and B are perpendicular to OA and OB respectively.
5--With angles BOA, OAD and OBD being 90º each, angle BDA must be 90º making OBDA a square with BD = AD.
6--Therefore, triangle ADB is isosceles with angle BDA = 90º.