Solving Congruent Triangles Inscribed in a Circle

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To solve for congruent triangles inscribed in a circle, the discussion highlights the importance of congruency criteria such as SSS, RHS, AAS, and SAS. The angles C and B are equal due to their position on the circumference, with chord A connecting them. The alternate angle theorem is applied to establish that angles PCB and PBC are equal, leading to the conclusion that sides PC and PB are equal, confirming SSS congruency. The angle between a tangent and a chord is also discussed, emphasizing its relevance in identifying congruency. Overall, the conversation revolves around applying geometric theorems and reasoning to demonstrate triangle congruence effectively.
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Sorry, it's upside down :(

How do i go about solving A. I know that for somthing to be congruent it needs to

SSS
RHS
AAS
SAS

I know that C = B because they lie on a circuference and a chord that binds them is A. Where do I go from now?

Thx
 
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thomas49th said:
I know that for somthing to be congruent it needs to

SSS
RHS
AAS
SAS

What does this mean?
 
abreivations of triangle congruency

take a look:
http://www.bbc.co.uk/schools/gcsebitesize/maths/shapeh/congruencyandsimilarityrev2.shtml
 
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Angles pcb and pbc are also equal (alternate angle theorem). Therefore, pc=pb (sides opp equal angles). This gives you two sides equal and 1 side common so sss congruency is established.
 
I don't see how it's alternate segment theory...

The angle between a tangent and a chord is equal to the angle made by that chord in the alternate segment.

I can't see it. Do you know a good technique for spotting it?

Thx
 
If you extend pc to some point, say q. Then angle acq is equal to angle abc. Apply the same thing on the other side of the quad.
 
OKay, this is my thinking
statement 1: AP=AP Reason: common line (S)
statement 2: <ABC = <ACB reason: given
statement 3: <PBC = <PCB reason: tangent from the same point P
statement 4: therefore <ABP = <ACP reason: see statement 2&3 (A)
statement 5: <CPA = <BPA reason: tangent cords are from the same point P (A)
statement 6: triangle ABP = triangle ACP reason: AAS

but I'm not sure for statement 5, as the diagram doesn't indicate anything...
 
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