Hi there.
I have an image of my question here. Here is an image with the colours crudely removed in case I am wrong and they are misleading. Here is the description.
Consider the triangle ABC inscribed in a circle. Extend CA to some point D. Take the angle bisector of DAB. It will intersect the circle at some point, P (you can assume this point exists and that the angle bisector is not tangent to the circle). Prove that PB = PC.
I've tried some proofs but so far I have actually managed to somehow prove that it could not possibly be the case. I am sure I am doing something wrong, because the conclusion does not make sense with the image.
If PB = PC, then BPC would be an isosceles triangle and its base angles would be equal.
Its base angles are PCB, PBC.
PCB = PCA + ACB.
Since PAB also subtends PB, and PAB = PAC + CAB,
PAC + CAB = PCA + ACB.
But the other base angle, PBC, is equal to PAC (subtends the same arc, PC).
So PCB can't equal PBC, since PBC = PAC + CAB while PBC = PAC (unless CAB is zero).
I know that the conclusion is wrong; it does not make sense and the question asks to prove that it is the case. I keep trying different methods to prove PC=PB and I'm just not seeing it...
EDIT// The above is actually kind of a roundabout way of doing it, sorry. You can just say that PC is subtended by PAC, while PB is subtended by PAC + CAB, so unless CAB is zero the two cannot be equal? (i.e., two chords of equal lengh = congruent arcs. Do congruent arcs have to be subtended by the same angle? I feel like I'm misunderstanding something pretty profoundly at this point...)
Anyway, I've tried doing it with sums of interior angles, using ABCP as a quadrilateral, proving PCB is isosceles, looking for congruent triangles... I can't see where to go
I have an image of my question here. Here is an image with the colours crudely removed in case I am wrong and they are misleading. Here is the description.
Consider the triangle ABC inscribed in a circle. Extend CA to some point D. Take the angle bisector of DAB. It will intersect the circle at some point, P (you can assume this point exists and that the angle bisector is not tangent to the circle). Prove that PB = PC.
I've tried some proofs but so far I have actually managed to somehow prove that it could not possibly be the case. I am sure I am doing something wrong, because the conclusion does not make sense with the image.
If PB = PC, then BPC would be an isosceles triangle and its base angles would be equal.
Its base angles are PCB, PBC.
PCB = PCA + ACB.
Since PAB also subtends PB, and PAB = PAC + CAB,
PAC + CAB = PCA + ACB.
But the other base angle, PBC, is equal to PAC (subtends the same arc, PC).
So PCB can't equal PBC, since PBC = PAC + CAB while PBC = PAC (unless CAB is zero).
I know that the conclusion is wrong; it does not make sense and the question asks to prove that it is the case. I keep trying different methods to prove PC=PB and I'm just not seeing it...
EDIT// The above is actually kind of a roundabout way of doing it, sorry. You can just say that PC is subtended by PAC, while PB is subtended by PAC + CAB, so unless CAB is zero the two cannot be equal? (i.e., two chords of equal lengh = congruent arcs. Do congruent arcs have to be subtended by the same angle? I feel like I'm misunderstanding something pretty profoundly at this point...)
Anyway, I've tried doing it with sums of interior angles, using ABCP as a quadrilateral, proving PCB is isosceles, looking for congruent triangles... I can't see where to go
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