Any solution?

Wait, what is $Q$?

sorry,typo corrected.thank you

in the 1st condition, do you mean by directed angles?

How can difference of two angles be equal to length of arc?

think of measures, subtracting degrees you get degrees

MRF2017 wrote: Given a circle and Points $P,B,A$ on it.Point $Q$ is Interior of this circle such that: $1)$ $\angle PAQ=90$. $ 2)PQ=BQ$. Prove that $\angle AQB - \angle PQA=\stackrel{\frown}{AB}$. proposed by Davoud Vakili, Iran. I think it is $\angle AQB+\angle PQA=\stackrel{\frown}{AB}$

Nope, it's minus.

Math-wiz wrote: Nope, it's minus. $\angle AQB-\angle PQA=\angle PQB.$ Now, tell me how come $\angle PQB=\angle AOB=\overset\frown {AB}?$

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I think @above is correct Math-wiz ...

I am weak at Geo and can't understand your solution. I just verified the problem from original source. Here's the official solution

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In the original sourse the problem starts by: In the figure below... So, it ought to have a figure, which was not given here. Obviously, different figures give different results. The fault is of MRF2017!

DEFINITIONS Call the circle $\omega$, and let it`s center be $O$. Let the extension of $AQ$ intersect $\omega$ for the second time at $C$. Draw segments $\overline{BC}$, $\overline{PB}$, and diameter $\overline{PC}$, obviously $O$ lies on $\overline{PC}$. Also let $\overline{PB}$ intersect $\overline{AQ}$ at $E$, and finally let $M$ be the foot of perpendicular from $O$ to $\overline{PB}$. PROOF First note that $M$ is midpoint of $\overline{PB} \implies M, Q,$ and $O$ are collinear Furthermore $\overline{MO}$ is midline of $\triangle{BPC}$, so $$\overline{MO}\parallel \overline{BC}\implies \angle{MQE}=\angle{BCA} \qquad (1)$$ Then we are left only with angle-chasing: First we see that \begin{align*}\angle{MQB} &= \angle{MQP} \\&=\angle{MQE}+\angle{PQA} \qquad (2)\end{align*}then \begin{align*}\angle{AQB}-\angle{PQA} &= (\angle{MQB}+\angle{MQE})-\angle{PQA} \\&=(\angle{MQE}+\angle{PQA}+\angle{MQE})-\angle{PQA} \qquad From (2) \\&= 2\angle{MQE} \\&= 2\angle{BCA} \qquad From (1) \\&=\angle{BOA} \\&=\stackrel{\frown}{AB}\end{align*}And we are done. $\blacksquare$

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No extensions. Nothing tricky Solution. Let $\angle QPB=\angle QBP=x$. Then $$\angle AQB - \angle PQA = \angle PQB - 2\angle PQA = (\pi-2x)-2(\pi-(90+x+\stackrel{\frown}{\frac{AB}{2}}))=\overarc{AB}$$

Let C be midpoint of PB. ∠QAP = 90 = ∠QCP ---> QCAP is cyclic. ∠AQB - ∠PQA = ∠BQC + ∠CQA - ∠PQA = ∠CQP + ∠CQA - ∠PQA = 2∠CQA = 2∠BPA = arc AB