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Thread: Proving Congruence

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    Member GAdams's Avatar
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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
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    A riddle wrapped in an enigma
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    Quote Originally Posted by GAdams View Post
    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
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  3. #3
    Member GAdams's Avatar
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    than you!
    Not sure how to find angle ABC...

    I can see that angle BPA and angle CPA are the same
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    A riddle wrapped in an enigma
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    Quote Originally Posted by GAdams View Post
    than you!
    Not sure how to find angle ABC...

    I can see that angle BPA and angle CPA are the same
    $\displaystyle m \angle BPA=10$

    $\displaystyle \triangle BPC$ is isosceles.

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

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

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

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

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