Question regarding inscribed angle theorem

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Sjein

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Is the angle PNQ=1/2POQ? They are subtended by the same arc.
But the problem appears when we use properties of a cyclic quadrilateral, namely sum of pair of opposite angles is equal to 180°. The angle opposite to PNQ is located to the right of the centre of the circle and must be equal (180-54)°=126°. However, it must be equal 54° because it is subtended by the same arc as angle POQ
 
pka said:
\(n^o=\frac{1}{2}\left(360^o-m(\widehat{PNQ)}\right)\)
Click to expand...
I am sorry but I do not understand your point.
I am asking to help me to find mistakes in my reasoning if there are any.
 
Suppose that we add a label \(R\) so that \(QRP\) is a triangle.
Then one-half the measure of the major arc \(\widehat{QRP}\) is equal to \(n^o=m(\angle PNQ)\)
 
[MATH][/MATH]
pka said:
Suppose that we add a label \(R\) so that \(QRP\) is a triangle.
Then one-half the measure of the major arc \(\widehat{QRP}\) is equal to \(n^o=m(\angle PNQ)\)
Click to expand...
Can you say if there are any errors in my reasoning?
 
Sjein said:
View attachment 20497
Is the angle PNQ=1/2POQ? They are subtended by the same arc.
But the problem appears when we use properties of a cyclic quadrilateral, namely sum of pair of opposite angles is equal to 180°. The angle opposite to PNQ is located to the right of the centre of the circle and must be equal (180-54)°=126°. However, it must be equal 54° because it is subtended by the same arc as angle POQ
Click to expand...
The error in your reasoning is that you used the angle POQ on the left side of O, 108, rather than the angle POQ on the right side of O, 360° - 108° = 252°, which is the central angle for the arc subtended by angle PNQ. Half of 252° is 126°.
 
Dr.Peterson said:
The error in your reasoning is that you used the angle POQ on the left side of O, 108, rather than the angle POQ on the right side of O, 360° - 108° = 252°, which is the central angle for the arc subtended by angle PNQ. Half of 252° is 126°.
Click to expand...
Why do you think that I used the wrong angle?
 
Last edited by a moderator:
Maybe I'm misreading what you claim is a problem; you skipped some steps, forcing me to make assumptions.

Please restate the OP, making it as clear as possible what you see as a conflict, and how you got there. If we see every step you took, we can point out which one is wrong.
 

Dr.Peterson said:
Maybe I'm misreading what you claim is a problem; you skipped some steps, forcing me to make assumptions.

Please restate the OP, making it as clear as possible what you see as a conflict, and how you got there. If we see every step you took, we can point out which one is wrong.
Click to expand...
I don't think that I missed any steps. But if you noticed that I had missed some, you are welcome to point out
 
Sjein said:
View attachment 20497
Is the angle PNQ=1/2POQ? They are subtended by the same arc.
But the problem appears when we use properties of a cyclic quadrilateral, namely sum of pair of opposite angles is equal to 180°. The angle opposite to PNQ is located to the right of the centre of the circle and must be equal (180-54)°=126°. However, it must be equal 54° because it is subtended by the same arc as angle POQ
Click to expand...
Where did that come from?

1595096483768.png
 
Sjein said:
I don't think that I missed any steps. But if you noticed that I had missed some, you are welcome to point out
Click to expand...
It's not necessarily that you yourself missed a step, but that you didn't tell us every step you did. We can't help without seeing the details of your thinking. It may also be helpful to label the rightmost vertex R so you can explicitly state what each angle you refer to is.
 
Sjein said:
Can you say if there are any errors in my reasoning?
Click to expand...
I agree with Prof Peterson that your arguments are hard to follow. My chief concern is for you to see this reasoning.
The measure of the arc \(\widehat{PNQ}=108^o\) because its central angle \(\angle POQ\) measures \(108^o\).
Hence, its complement arc \(\widehat{PRQ}=252^o\) that is subtended by \(\angle PNQ\) thus \(n^0=\frac{1}{2}(252^o)=126^o\).
It not what you did incorrectly but a different approach. That was my concern.
 
Dr.Peterson said:
The error in your reasoning is that you used the angle POQ on the left side of O, 108, rather than the angle POQ on the right side of O, 360° - 108° = 252°, which is the central angle for the arc subtended by angle PNQ. Half of 252° is 126°.
Click to expand...
Now I understand my mistake. Thanks to all of you for help
 
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