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
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
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
$\displaystyle m \angle BPA=10$
$\displaystyle \triangle BPC$ is isosceles.
$\displaystyle m \angle PBC=m \angle PCB=80$
$\displaystyle m \widehat{BC}=160$
$\displaystyle m \widehat{BAC}=200$
$\displaystyle m \widehat {AB}=m \widehat {AC}=100$
$\displaystyle m \angle ABC=\frac{1}{2} m \widehat{AB}=50$