# Some angles of triangle inscribed in circle and intersection of its bisectors with that circle

A $\triangle ABC$ has been inscribed in a circle. The bisectors of angles $A$, $B$ and $C$ meet the circle at $P$, $Q$ and $R$ respectively.

If $\angle BAC = 50^\circ$, then what would be the value of $\angle QPR$, $\angle BQP$, $\angle ABC$, $\angle ACB$, $\angle QRC$, $\angle RPQ$ and $\angle QRP$ ?

• Note that we definitely cannot determine $\angle ABC$ and $\angle ACB$ from the information given. – André Nicolas Oct 21 '13 at 6:34
• FTRE questions, isn't it? But you need to rephrase your questions. – Sawarnik Oct 25 '13 at 16:58

I'll describe my methods:

1: First of all $∠BAC = 50^\circ$. Given that $AP$ is the angle bisector of $∠A$, we conclude that $∠BAP = 25^\circ$ and $∠CAP=25^\circ$.

2:We know that in a circle the angles formed by a chord on the circumference equal one another (proof in the image here: http://en.wikipedia.org/wiki/Inscribed_angle#Theorem).

Therefore, considering chord $BP$, we have $∠BAP=∠BRP$, or $∠BRP=25^\circ$. Same way, considering chord $PC$, we have $∠CQP = 25^\circ$. With the same methods, we get $∠CQR = \frac{∠B}{2}$ and $∠BRQ = \frac{∠C}{2}$.

3: Doing this with $∠QPR$, we get $∠APQ =\frac{∠C}{2}$ and $∠APR = \frac{∠B}{2}$. So, $$∠QPR = \frac{∠B}{2} + \frac{∠C}{2} = \frac{∠B+∠C}{2} = \frac{180^\circ-∠A}{2} = 90^\circ - \frac{∠A}{2} = 65^\circ$$

Similarly, we have $∠PQR= 90^\circ -\frac{∠C}{2}$ and $∠PRQ = 90^\circ-\frac{∠B}{2}$.

At this point it would be good to realize that we cannot calculate all the angles you asked for, however, as we saw, we can calculate it at least in terms of other angles. We can definitely not calculate $∠C$ and $∠B$ from the information, as Nicolas said in the comments. The only restriction which lies on them is that $∠B+∠C = 130^\circ$ [angle-sum property], which can be used to calculate the other when one is given. And if we are given one of them, you could substitute them into the formulas we derived, to obtain the results.

4: Now, $∠QBP = ∠QRP = 90^\circ - \frac{∠B}{2}$, $∠BQP =∠PAB =25^\circ$, $∠BPQ = ∠BCQ = \frac{∠C}{2}$. Remember, the rule I told in the 2nd point. These follow directly from it. Using that rule, you can calculate any other angle you want.

see that $\angle$QPR=$\angle$QPA+$\angle$RPA

• $\angle$QPA=$\angle$QBA=$\frac{B}{2}$
• SIMILARLY $\angle$RPA=$\frac{C}{2}$
• so $\angle$QPR=$\frac{B}{2}$+$\frac{C}{2}$=$\frac{1}{2}$(180-50)=65
• $\angle$BQP=$\frac{A}{2}$=25
• but there is no sufficient information to find the angles ABC;ACB;QRC;RPQ;QRP