Solving Congruent Triangles Inscribed in a Circle

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SUMMARY

This discussion focuses on solving congruent triangles inscribed in a circle using various congruency criteria: SSS (Side-Side-Side), RHS (Right angle-Hypotenuse-Side), AAS (Angle-Angle-Side), and SAS (Side-Angle-Side). The user establishes that angles C and B are equal due to their positions on the circumference, leading to the conclusion that triangle ABP is congruent to triangle ACP using the AAS criterion. The alternate angle theorem and the properties of tangents and chords are also discussed as essential concepts in proving congruency.

PREREQUISITES
  • Understanding of triangle congruency criteria: SSS, RHS, AAS, SAS
  • Knowledge of the alternate angle theorem
  • Familiarity with properties of tangents and chords in circles
  • Ability to interpret geometric diagrams and statements
NEXT STEPS
  • Study the properties of inscribed angles and their relationships in circles
  • Learn about the alternate segment theorem and its applications
  • Practice solving problems involving triangle congruency in circle geometry
  • Explore advanced geometric proofs involving tangents and chords
USEFUL FOR

Students studying geometry, mathematics educators, and anyone interested in mastering the concepts of triangle congruency and circle theorems.

thomas49th
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Question003.jpg


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
 
Last edited by a moderator:
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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