Proving Congruence

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April 20th 2009, 09:14 AM
GAdams

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Proving 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
April 20th 2009, 09:35 AM
masters

Quote:

Originally Posted by GAdams

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 $\triangle APB \cong \triangle APC$

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

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

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

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

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

$\triangle APB \cong \triangle APC$ - SSS Postulate
April 20th 2009, 10:03 AM
GAdams

than you!
Not sure how to find angle ABC...

I can see that angle BPA and angle CPA are the same
April 20th 2009, 12:51 PM
masters

1 Attachment(s)

Quote:

Originally Posted by GAdams

than you!
Not sure how to find angle ABC...

I can see that angle BPA and angle CPA are the same

$m \angle BPA=10$

$\triangle BPC$ is isosceles.

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

$m \widehat{BC}=160$

$m \widehat{BAC}=200$

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

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