File attached. I'm not too sure how to do this, here's what I have so far:

BP = PC (tangents that cross coming from a circle) [side]

angle ACX = angle ABC (alternate segments)

angle YBA = angle ACD (alternate angles)

Therefore, angle ABC = ACB

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- Apr 20th 2009, 09:14 AMGAdamsProving Congruence
File attached. I'm not too sure how to do this, here's what I have so far:

BP = PC (tangents that cross coming from a circle) [side]

angle ACX = angle ABC (alternate segments)

angle YBA = angle ACD (alternate angles)

Therefore, angle ABC = ACB - Apr 20th 2009, 09:35 AMmasters
Hi GAdams,

Prove $\displaystyle \triangle APB \cong \triangle APC$

$\displaystyle \angle ABC \cong \angle ACB$ - Given

$\displaystyle \overline{PB} \ \ and \ \ \overline{PC}$ are tangent to the circle - Given

$\displaystyle \overline{AB} \cong \overline{AC}$ - If two angles of a triangle are congruent, then the sides opposite these angles are congruent.

$\displaystyle \overline{BP} \cong \overline{CP}$ - If two segments from the same exterior point are tangent to a circle, then they are congruent.

$\displaystyle \overline{AP} \cong \overline {AP}$ - Reflexive property of congruence.

$\displaystyle \triangle APB \cong \triangle APC$ - SSS Postulate - Apr 20th 2009, 10:03 AMGAdams
than you!

Not sure how to find angle ABC...

I can see that angle BPA and angle CPA are the same - Apr 20th 2009, 12:51 PMmasters
$\displaystyle m \angle BPA=10$

$\displaystyle \triangle BPC$ is isosceles.

$\displaystyle m \angle PBC=m \angle PCB=80$

$\displaystyle m \widehat{BC}=160$

$\displaystyle m \widehat{BAC}=200$

$\displaystyle m \widehat {AB}=m \widehat {AC}=100$

$\displaystyle m \angle ABC=\frac{1}{2} m \widehat{AB}=50$