ABC is a triangle inscribed within a circle with center O, CM⊥AB M∈AB, BN⊥AC N∈AC. Prove AO⊥MN.

[frosty sail]

Please help me with this question.

[whole ermine]

...there's no AO. O is the circle. You can't have a line between a point and a circle.

[frosty sail]

I mean O is the center of the circle, (O). Sorry for misunderstanding

[whole ermine]

So... draw the picture?

[frosty sail]

yes please wait a minute

[whole ermine]

Actually, this is false, because AO doesn't even necessarily intersect MN.

[frosty sail]

Can you explain why please?

[whole ermine]

,rotate

[frosty sail]

A, B, C is a point on the circle, sorry I couldn't find the correct term to say that.

[whole ermine]

Inscribed. Triangle ABC is inscribed within a circle with center O.

[frosty sail]

thanks

Here's a picture for the question

[whole ermine]

Also, shouldn't M be on AB?

[frosty sail]

yes it should

[whole ermine]

You said it was on BC.

[frosty sail]

oh I apologise for making so much mistake and the confusion it caused

[brazen canopy]

from the question so it means that B = M ?

[whole ermine]

...no, the question was incorrectly written.

[frosty sail]

ABC is a triangle intesected circle with center O, CM⊥AB M∈AB, BN⊥AC N∈AC. Prove AO⊥MN.

[whole ermine]

Inscribed within a circle.

[frosty sail]

ABC is a triangle inscribed circle with center O, CM⊥AB M∈AB, BN⊥AC N∈AC. Prove AO⊥MN.

[whole ermine]

Inscribed within a circle.

[frosty sail]

ABC is a triangle inscribed within a circle with center O, CM⊥AB M∈AB, BN⊥AC N∈AC. Prove AO⊥MN.

[whole ermine]

It's very important in math to say exactly what you mean.

[frosty sail]

Thank you for your patient and your correcting my mistake

[whole ermine]

So first we should note that each of the lines CM and BN split our triangle into two right triangles, right?

[frosty sail]

yes Techie Literate (I don't know how to address you)

[whole ermine]

And in fact we know that CM and BN must intersect, right?

[frosty sail]

yes

[whole ermine]

Hmm.

Honestly, I'm kind of just making observations as I notice them.

But I don't actually know a whole lot about incircles and circumcircles of triangles.

[frosty sail]

It's fine, thank you for helping me renaming my post.

This is the picture in case someone need it.

[frosty sail]

@brazen canopy

Hi thanks for responding

[paper dome]

angle chasing

Note that $\angle BMC = \angle BNC = 90^\circ$

[silver halo]

k12byda5h

[paper dome]

so $B,C,M,N$ are concyclic

[silver halo]

k12byda5h

[paper dome]

Then, you can chase that $\angle BAO = 90^\circ - \angle ACB = \angle AMN$

[silver halo]

k12byda5h

[paper dome]

so we finish

[frosty sail]

what does concyclic mean?

[paper dome]

It lies on a same circle

[frosty sail]

Can you explain how BAO = 90 - ACB = AMN please?

[paper dome]

Do you understand this now?

[frosty sail]

No I dont

[paper dome]

Do you know these things?

[frosty sail]

I don't

I'm in grade 9 and they haven't teached me these

[paper dome]

It's hard to solve the problem without this knowledge

so how does your teacher give you this prob? If I understand the teacher's intention, I might figure out the expected solution

[frosty sail]

well I was learning some theorem about circle

do you need the list of theorem in that lesson?

[paper dome]

here

I mean,

that is necessary

you can give me your list if you want

[frosty sail]

I will but it will take some minutes to translate them

[frosty sail]

M belongs to the circle O, so OM = R
The circle is the intersection of the three perpendicular bisectors of the triangle.

This part is second picture in right column (The center of the circumcircle of a right triangle is the midpoint of the diameter. In a triangle has one side that is the diameter of the circumcircle, then that triangle is a right triangle)

In a circle, a diameter perpendicular to a string passes through the midpoint of that string
The diameter passing through the midpoint of a string not through the center is perpendicular to that string

This part is included in the finale part of your picture (If a line is tangent to a circle, then it is perpendicular to the radius passing through the point of contact
If a line passes through a point of a circle and is perpendicular to the radius passing through that point, then the line is tangent to the circle.
If two tangents to a circle intersect at a point, then that point is:

• Equidistant from the two points of contact,
• The ray from the center passing through that point is the bisector of the angle formed by the two radius passing through the points of contact,
• The ray from that point passing through the center is the bisector of the angle formed by the two tangents.)

The center of the circumcircle of the triangle is the intersection of the bisectors of the triangle.
The center of the circumcircle of a triangle (I mean the circle tangent to one side of the triangle and tangent to the extensions of the other two sides) is the intersection of the bisectors of the two exterior angles of a triangle or the intersection of the bisector of the interior angle and the bisector of the non-adjacent exterior angle.
I use google translate so some of them might be incorrectly translated.

[paper dome]

1. M belongs to the circle O, so OM = R
2. The center of circle is the intersection of the three perpendicular bisectors of the sides of triangle (actually, the perpendicular bisector of any chord of a circle passes through its center).
3. The center of the circumcircle of a right triangle is the midpoint of the hypotenuse
4. (converse of 3) In a triangle has one side that is the diameter of the circumcircle, then that triangle is a right triangle
5. In a circle, a diameter perpendicular to a chord passes through the midpoint of that chord
6. (converse of 4) The diameter passing through the midpoint of a chord not a diameter is perpendicular to that chord
7. If a line is tangent to a circle, then it is perpendicular to the radius passing through the point of contact
8. (converse of 7) If a line passes through a point of a circle and is perpendicular to the radius passing through that point, then the line is tangent to the circle.
9. If two tangents to a circle intersect at a point, then that point is:

• Equidistant from the two points of contact,
• The ray from the center passing through that point is the bisector of the angle formed by the two radius passing through the points of contact,
• The ray from that point passing through the center is the bisector of the angle formed by the two tangents.

9. (same as 2.) The center of the circumcircle of the triangle is the intersection of the bisectors of the triangle.
10. The center of the circumcircle of a triangle (I mean the circle tangent to one side of the triangle and tangent to the extensions of the other two sides) is the intersection of the bisectors of the two exterior angles of a triangle or the intersection of the bisector of the interior angle and the bisector of the non-adjacent exterior angle.

lemme teach you a theorem related to circle.

btw, do you know about angle chasing?

[frosty sail]

Number 9 is refering to a circle smaller than triangle and number 2 is refering to a circle larger than the triangle (I mean the triangle inscribed within a circle) and thank you for teaching english vocab

[paper dome]

like, sum of angles in a triangle is 180

[frosty sail]

yes

[paper dome]

ok

By this picture, we get that $2 \angle BAC = \angle BOC$

[silver halo]

k12byda5h

[paper dome]

oh sry sry, in that case, it is called "incircle". The circle you mentioned in 10) is callled "excircle"

[paper dome]

Do you wanna prove?

[frosty sail]

I regret to say that I have to go now, I will be back in 1 hour, if you can't answer my question then, please write it so I can see it later, thank you for your patient and kindness.

[paper dome]

ah okay

[paper dome]

So, let's prove this theorem first.

Draw $OA$. You know that the length of $OA = OB = OC$. Hence, $\angle BAO = \angle ABO$ and $\angle ACO = \angle OAC$. Thus, $\angle BOC = \angle OBA+ \angle OCA + \angle BAC$. Therefore $\angle BOC = 2 \cdot \angle BAC$ as desired

[silver halo]

k12byda5h

[paper dome]

now, see this picture. By the previous theorem we proved, we get that $\angle BOC = 2 \angle BAC$. Similarly, $\angle BOC = 2 \angle BDC$. Thus, $\angle BAC = \angle BDC$

[silver halo]

k12byda5h

[paper dome]

This is really important. Keep that in mind

You said you don't know what concyclic is, right?

you know that for any 3 points, there is only one circle passing through them which is the circumcircle of the triangle forming by those 3 points.

However, it is not true that you can always find a circle that passes through 4 points.

But if there is, we call those 4 points concyclic.

eg. these points are concyclic

because they lie on the same circle

[paper dome]

Back to here. The statement is that "if A,B,C,D are concyclic, angle BAC = angle BDC"

Its converse is also true. if angle BAC = angle BDC, then A,B,C,D are concyclic

(for the reason, let's talk when you are here. There are sth that you have to concern with too. Tell me if you are here)

[paper dome]

oh cheese

I've just realized sth\

You can use 3) in your list to show that B,C,M,N are concyclic with a circle with diameter BC

AAAAAAAAAAAAAAAA

[frosty sail]

@paper dome Hello, I'm back, sorry for leaving suddenly. I have read all your message.

[paper dome]

What I showed is the picture 2 and 3 or this pic

[frosty sail]

well I understand 2 and 3 now

[paper dome]

how about 6?

nvm, I didn't explain

[frosty sail]

I dont

[paper dome]

ok I will prove that $\angle BAC + \angle BDC = 180^\circ$

[silver halo]

k12byda5h

[paper dome]

It is nearly the same as this one, but A,D lie on the different side of BC

[paper dome]

By what we proved:
2 angle BAC = pink angle
2 angle BDC = blue angle
2(angle BAC + angle BDC) = pink + blue =360

[paper dome]

Therefore, we get this one

Does it make sense? we will be ready for your original question

[frosty sail]

yes it does

I'm ready for my original question now

[paper dome]

place the pic here again

[paper dome]

By here, B,C,M,N are concyclic

[paper dome]

By here, angle BCA + angle BMN = 180

Therefore angle BCA = angle AMN

[paper dome]

Now, from here, 2angle ACB = angle AOB

since AO = BO, angle BAO = angle OBA

Hence, $\angle BAO = \frac{180- \angle AOB}{2} = \frac{180 - 2\angle ACB}{2} = 90^\circ - \angle ACB = 90^\circ - \angle AMN$

[silver halo]

k12byda5h

[paper dome]

Thus, $\angle MAO+\angle AMN = 90^\circ$

[silver halo]

k12byda5h

[paper dome]

This implies that MN is perpendicular to AO