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  1. Triangle ABC is inscribed in a circle, and the bisectors of the angles meet the circumference at XYZ. Show that the angles of the triangle XYZ are respectively $90^\circ- A/2$, $90^\circ- B/2$, $90^\circ- C/2$.

  2. AB, AC are two chords of the circle. P, Q are midpoints of minor arcs cut off by them. If PQ is joined, cutting AB at X, and AC at Y. Show that AX = AY.

  3. Two equal circles intersect at A, B. A straight line PAQ is drawn, terminated by the circumference. Show that BP = BQ.

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It is helpful if you can ask one question at a time. Also it would be useful if you could explain what makes these questions difficult and how you have tried to approach them. Have you drawn a diagram? – Mark Bennet Sep 7 '13 at 17:29
@Sawarnik, Welcome to MSE! Please keep in mind that you need to show what you have tried in your problem in order to get help you ask for! – math Sep 7 '13 at 17:29
These questions were in a course book, with which I practice, and are the only ones that could not be solved. I have put in all my efforts but have largely been of no avail. Of course, I have drawn diagrams, but they seem to lead nowhere. Please help. – Sawarnik Sep 7 '13 at 17:35
In the first question, I can see something but could not convert it to a result. As you can see, 90 - a/2 is the angle direct below a, on the center of bisector of angles. But I am unable to prove that those two angles are congruent. – Sawarnik Sep 7 '13 at 17:38
The first question is simple angle chasing. Use basic circle properties like angle in same arc. WOuld be helpful if you knew that the angle bisectors pass through the same point too. – Calvin Lin Sep 7 '13 at 17:40
up vote 2 down vote accepted

For question 1, it's immediate angle chasing. Just do it.

$\angle ZXY = \angle ZXA + \angle YXA = \angle ZCA + \angle YBA = ?? + ?? = 90 - \frac{\alpha}{2}$

For question 2, you want to use the fact that if the angle subtended by 2 chords is dependent only on the angle measure between the pairs of endpoints. In fact, if $W, X, Y, Z$ are points on a circle, then the angle between $WY$ and $XZ$ is half of the angle measure $\arc{WX} + \arc{ YZ}$. use this to show that $AXY$ is an isosceles triangle.

For question 3, since the circles are equal, the angle subtended by $AB$ at $P$ is equal to the angle subtended by $AB$ (of the same length) at $Q$. Hence $BPQ$ is an isosceles triangle.

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Thanks, I am done on 3, looking into other answers. – Sawarnik Sep 7 '13 at 18:00
@CalvinLin Can you please explain the lemma that you've used in question 2, a little bit better? A drawing would do better. – Stefan4024 Sep 8 '13 at 16:15

For #1, remember the basic fact about the arc subtended by an inscribed angle. $\angle X$ subtends half the arc of $\angle B$ plus half the arc of $\angle C$.

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