Angles inscribed in circles part 1

[grimTesseract]

Homework Statement

OK, I am new to these kinds of problems and I am trying to learn the appropriate properties but they are proving somewhat difficult for me... I hope I am going in the right direction.

Homework Equations

[/B]
The first problem corresponds to the figure with 'Rep' in the filename. I need merely to measure the angle 'a.' I'm a little stuck because, given the angle 105 degrees, I was under the impression that the angles of the opposite vertices of the triangle must sum 105, and that the adjacent angle must be 180 - 105 = 75, but in the diagram on my worksheet the triangle looks equilateral.

The second problem will correspond to the uploaded figure with 'exam' in the filename.
Given a circumference with center $O$, radius $r$ and diameter $AB$, we consider the point $E$, in the arc $AB$ given the figure. Let $M$ be a variable point on the arc $AE$, and on the line containing the segment $MA$ let $C$ be the point such that $MC = MB$. $MF$ is the perpendicular line to $CB$ through $M$, $F$ being the intersection of the line and the circumference. $MF \cap CB = {H}$

a) Classify the triangle MBC by its sides and angles.
b) Prove that the angle FMB is constant when varying M, and thus that F is a fixed point.
c) What geometric figure is included in the locus of the point H when M varies? (CHM = 90 degrees)
d) Prove that the triangles BHO and BCA are similar
e) Utilizing parts (c) and (d), deduce the geometric figure included in the locus of C.

I am significantly more lost here. (c) and logically (e) confound me and I'm not sure how to go about solving them. I can't tell how to deduce the angles of MBC but I would say it's isosceles, given that I'm told $MC = MB$. Proving the part (b) seems like it must somehow relate to the central angle, but I don't see exactly how I must prove this. Finally, I can see that BHO is a scaled down version of BCA, but I don't know how to prove the similarity in angles.

All help is appreciated here.

 

[Mark44]

Homework Statement

OK, I am new to these kinds of problems and I am trying to learn the appropriate properties but they are proving somewhat difficult for me... I hope I am going in the right direction.

Homework Equations

[/B]
The first problem corresponds to the figure with 'Rep' in the filename. I need merely to measure the angle 'a.' I'm a little stuck because, given the angle 105 degrees, I was under the impression that the angles of the opposite vertices of the triangle must sum 105, and that the adjacent angle must be 180 - 105 = 75, but in the diagram on my worksheet the triangle looks equilateral.

In the first problem, you're missing the fact that triangle AOB is isosceles, so the base angles have to be equal. "Looks equilateral" doesn't mean that the triangle actually is equilateral.

The second problem will correspond to the uploaded figure with 'exam' in the filename.
Given a circumference with center $O$, radius $r$ and diameter $AB$, we consider the point $E$, in the arc $AB$ given the figure. Let $M$ be a variable point on the arc $AE$, and on the line containing the segment $MA$ let $C$ be the point such that $MC = MB$. $MF$ is the perpendicular line to $CB$ through $M$, $F$ being the intersection of the line and the circumference. $MF \cap CB = {H}$

a) Classify the triangle MBC by its sides and angles.
b) Prove that the angle FMB is constant when varying M, and thus that F is a fixed point.
c) What geometric figure is included in the locus of the point H when M varies? (CHM = 90 degrees)
d) Prove that the triangles BHO and BCA are similar
e) Utilizing parts (c) and (d), deduce the geometric figure included in the locus of C.

I am significantly more lost here. (c) and logically (e) confound me and I'm not sure how to go about solving them. I can't tell how to deduce the angles of MBC but I would say it's isosceles, given that I'm told $MC = MB$. Proving the part (b) seems like it must somehow relate to the central angle, but I don't see exactly how I must prove this. Finally, I can see that BHO is a scaled down version of BCA, but I don't know how to prove the similarity in angles.

All help is appreciated here.

 

[grimTesseract]

In the first problem, you're missing the fact that triangle AOB is isosceles, so the base angles have to be equal. "Looks equilateral" doesn't mean that the triangle actually is equilateral.

I understand! So the adjacent angle to the exterior angle I'm given must be 180 - 105 = 75, and the other two must be 105/2 = 52.5. Thus a must be 52.5 + 75 = 127.5

 

[Mark44]

I understand! So the adjacent angle to the exterior angle I'm given must be 180 - 105 = 75, and the other two must be 105/2 = 52.5. Thus a must be 52.5 + 75 = 127.5

Right.

 

[Mark44]

c) What geometric figure is included in the locus of the point H when M varies? (CHM = 90 degrees)
d) Prove that the triangles BHO and BCA are similar
e) Utilizing parts (c) and (d), deduce the geometric figure included in the locus of C.

For part c), I'm assuming that point B is fixed. Point F has to move in order to keep the angle at point H a right angle. I don't know what path point H moves along, but here is something I would try.
1) Draw four or five figures with line segment CB in different orientations, with M in various positions between A and E, to see if you can get an idea of the locus of point H. Make the drawings as faithful as you can to the conditions given in the problem. I'm guessing that point H moves along an arc of a circle, but I'm not certain of this.

For part e), again assuming point B is fixed, if triangles BHO and BCA are similar (since you didn't ask about it, I'm assuming you were able to prove this), then ACHO is a trapezoid, with OH and AC being parallel, independent of where M is chosen. What happens if M is exactly at A? What happens if E is right at B, and M is also at B?
Note: the paragraph above was edited to say "trapezoid" instead of "parallelogram."

 

[grimTesseract]

I will reply with my attempt at the rest of the problem shortly, but regarding the similarity of triangles BHO and BCA:

H and C are both on the same line, as A and O are on the parallel. Is this enough to prove that the triangles share two angles and must thus be similar?

 

[Mark44]

H and C are both on the same line, as A and O are on the parallel. Is this enough to prove that the triangles share two angles and must thus be similar?

Angle ABC is the same for both triangles, but I don't see how you get another angle in each triangle being equal.

Note that I edited my previous reply. I had mistakenly said that AOHC was a parallelogram. If you can show that the two triangles are similar, then it must be true that AC and OH are parallel.