Circle question (1 Viewer)

[ronnknee]

This is suppose to be a simple question but I can't see the way to get the answer

From Fitzpatrick

27. AB and AC are equal chords of a circle. AD and BE are parallel chords through A and B respectively. Prove that AE is parallel to CD

Basically:
- the blue lines are equal
- the orange lines are parallel
- prove the red lines are parallel

Note: BE is not necessarily a right angle and AE is not necessarily the diameter. They just happen to be like that when I drew it according to the conditions

 

[shaon0]

This is suppose to be a simple question but I can't see how the way to get the answer

From Fitzpatrick

27. AB and AC are equal chords of a circle. AD and BE are parallel chords through A and B respectively. Prove that AE is parallel to CD

Basically:
- the blue lines are equal
- the orange lines are parallel
- prove the red lines are parallel

Note: BE is not necessarily a right angle and AE is not necessarily the diameter. They just happen to be like that when I drew it according to the conditions

Something to do with angles..... Actually can you keep 'running' DC through the circle and setup a tangent to point A and prove AED(and external point to tangent) is a parallelogram?

....Sorry haven't done Year 12 3unit Maths yet

 

[ronnknee]

Is Triangle ABE isoceles? since AB and BE are perpendicular bisectors of one another

Not necessarily

 

[shaon0]

Not necessarily

Okay but that fact may not be necesary either way.

 

[lyounamu]

Ok, I proved that AE is a diameter. Now, what?

Let me get this straight first and then I will write up my solution. (extremely long)

 

[Iruka]

There is no reason why AE should be a diameter.

You have to join BC. Then use properties of cyclic quads (among other things) to show that angles CDA and DAE are supplementary.

 

[Aerath]

There is no reason why AE should be a diameter.

Yeah, except for the fact that he said he had proved that AE was the diameter.

 

[lyounamu]

There is no reason why AE should be a diameter.

You have to join BC. Then use properties of cyclic quads (among other things) to show that angles CDA and DAE are supplementary.

I know that. That's why I proved it. I joined BC and CE. Then I called the point where BC and AE intersects, O. Then I proved that AOB is 90 degrees, hence the line must be a diameter since the angle in a semicircle is a right angle.

 

[JSBboag]

This is suppose to be a simple question but I can't see the way to get the answer

From Fitzpatrick

27. AB and AC are equal chords of a circle. AD and BE are parallel chords through A and B respectively. Prove that AE is parallel to CD


Basically:
- the blue lines are equal
- the orange lines are parallel
- prove the red lines are parallel

Note: BE is not necessarily a right angle and AE is not necessarily the diameter. They just happen to be like that when I drew it according to the conditions

follow link --> it will show you that the lines have common gradients of 'x' due to corresponding angle and vertical angle.
Tell me if you can't access it or understand it?

View attachment 16688

 

[Iruka]

Yeah, except for the fact that he said he had proved that AE was the diameter.

No, that is not the case. If he proved that AE is a diameter then he is making some tacit assumptions that are not in the question.

Basically you can locate E anywhere on the circle you like, so long as you also move D so that AD and BE are parallel.

You can in fact draw the diagram with E lying in between A and B. (Try it and see!)

 

[lyounamu]

No, that is not the case. If he proved that AE is a diameter then he is making some tacit assumptions that are not in the question.

Basically you can locate E anywhere on the circle you like, so long as you also move D so that AD and BE are parallel.

You can in fact draw the diagram with E lying in between A and B. (Try it and see!)

Well, I don't think I made any assumption back then.

I just solved what I was given in the question. And it came up that the angle was 90 degrees which means that AE is a diameter.

 

[Iruka]

Try doing what I just suggested - draw the diagram with E in between A and B. Then you will see that the point of intersetion between BC and AE does not even lie within the circle.

 

[lyounamu]

Ok, here we go:

Construction: join BC and CE.

Call the point where the line BC and AE intersects O.

Angle CBE = A CAE = y (let's call this y) (angles in the same segment are equal)
A BCA = A AEB = x (let's call this x) (angles in the same segment are equal)
A BAE = y (because triangle AOC is congruent to the triangle AOB)
Therefore, angle BCE = y.
And A ABC = x and A AEC = x (angles in the same segment)

Now, you can see that all four triangles (i.e. triangle AOC, AOB, BOE and COE) are similar since they all have 2 equal angles. Therefore the other angle must be equal for all triangels. Let this angle be x. All these 4xs in the middle add up to 360, meaning that x=90. Therefore, AE is a diameter since the angle in a semicircle is 90 degrees.

Well, that's as far as I can go for now. I will come back and re-edit if I need to. I shall be off for a minute for a dinner.

 

[Iruka]

A BAE = y (because triangle AOC is congruent to the triangle AOB)

And there precisely is your tacit assumption. You haven't proved that these triangles ARE congruent.

We know that AB=AC and that OA is common to both triangles, but we need to know either the included angles, or the length of the other sides before we can say that they are congruent.

 

[lyounamu]

And there precisely is your tacit assumption. You haven't proved that these triangles ARE congruent.

We know that AB=AC and that OA is common to both triangles, but we need to know either the included angles, or the length of the other side before we can say that they are congruent.

Well, I didn't want to explain this as it will make me write more lines. Ok, I will show you here:

AB = AC
AO is a common side.
And A ACO = A ABO (base angles of an isoceles triangle are equal)
Therefore, they are congruent. (SAS)

Sorry for being lazy.

EDIT: Just realised that I could not use SAS since the angle given is not an included angle. I shall think of a new way... if I have time before I go to bed.

 

[Iruka]

You can't prove it as it is not necessarily true. Besides, you don't need to prove it:

Let angle ACB = x, as suggested by Iyounamu

Then angle ACB = angle CBA = x (base angles in isosceles triangle ACB)

Also, angle AEB = angle ACB = x (angles subtending the same segment)

Furthermore, angle DAE = AEB = x (alternating angles on parallel lines, AC and BE)

But because ABCE is a cyclic quadrilateral, angle CDA = 180 - angle ABC = 180 - x.

Consequently, angle CDA + angle DAE = 180-x + x = 180.

Thus DC is parallel to AE.

 

[ronnknee]

Ok, here we go:

Construction: join BC and CE.

Call the point where the line BC and AE intersects O.

Angle CBE = A CAE = y (let's call this y) (angles in the same segment are equal)
A BCA = A AEB = x (let's call this x) (angles in the same segment are equal)
A BAE = y (because triangle AOC is congruent to the triangle AOB)
Therefore, angle BCE = y.
And A ABC = x and A AEC = x (angles in the same segment)

Now, you can see that all four triangles (i.e. triangle AOC, AOB, BOE and COE) are similar since they all have 2 equal angles. Therefore the other angle must be equal for all triangels. Let this angle be x. All these 4xs in the middle add up to 360, meaning that x=90. Therefore, AE is a diameter since the angle in a semicircle is 90 degrees.

Well, that's as far as I can go for now. I will come back and re-edit if I need to. I shall be off for a minute for a dinner.

Wrong

O may not necessarily lie on AE, as ABE may not necessarily be a right angle in other cases

 

[lyounamu]

Wrong

O may not necessarily lie on AE, as ABE may not necessarily be a right angle in other cases

No, I didn't mean O as in origin. I just named it O. O is just the point where AE and BC intersects. From that, I tried proving that all the triangles are similar but I didn't get it right in the end.

 

[JSBboag]

i already proved that they are parallel with a much easier method.

 

[lyounamu]

i already proved that they are parallel with a much easier method.

Care to share with us?