Proving Equal Diagonals in a Quadrilateral Inscribed in a Circle

[nothing123]

1. A, B and C are points on the circumference of a circle, centre O, and ∠BAC=115°. Calculate the number of degrees in ∠OBC.

I drew a diagram but could not come up with anything primarily because ∠BAC and centre O lie on opposite sides of chord BC (draw a quick diagram if you don't know what I mean).

2. ABCD is a quadrilateral inscribed in a circle and AB = CD. Prove that AC = BD.

This is pretty simple I think, if AB = CD, then BC = DA (do I need to say why?). Then just use Pythag to prove the diagonals are equal.

 

[VietDao29]

1. A, B and C are points on the circumference of a circle, centre O, and ∠BAC=115°. Calculate the number of degrees in ∠OBC.

I drew a diagram but could not come up with anything primarily because ∠BAC and centre O lie on opposite sides of chord BC (draw a quick diagram if you don't know what I mean).

Hmm, I'll give you some hints to start off the problem.
Have you learned the relation between the central angle, and an inscribed subtended by the same chord, and on the same side of the chord? Hint: The chord is BC.

2. ABCD is a quadrilateral inscribed in a circle and AB = CD. Prove that AC = BD.

This is pretty simple I think, if AB = CD, then BC = DA (do I need to say why?). Then just use Pythag to prove the diagonals are equal.

Uhmm, this is not correct.
How can you go from AB = CD to BC = DA? And the problem does not state that there should be any right angle. How can you use Pythagorean Theorem?
------------------
If AB = CD, then what can you say about the two angles: ∠BDA, and ∠CAD?
Hint, you may want to use congruent triangles.
Can you go from here? :)

 

[AKG]

For #1, it doesn't matter that O is on the other side of the chord BC. You can still use the theorem to find some angle related to <BOC. With this, you can find the angle <BOC, and then use this, along with a key fact about the triangle OBC to find the angle <OBC.

For 2, use the cosine law, and some facts you know about the angles where 2 lines cross.

Also, can I ask what course you're doing this for? I remember doing this stuff well before calculus, i.e. grade 6.

 

[0rthodontist]

For #2 prove the fact that ABD is congruent to ACD.

 

[nothing123]

For #1, I still can't grasp how the Angles at the Circumference Property is applicable in this problem because the central angle and ∠BAC do not stand on the same arc (or chord).

AKG, I'm doing this for Grade 12 Discrete Math where we just started learning about circles. I can understand how you may think this is very basic (and it probably is), I just haven't fully understood the concepts and have not practiced enough to quickly recognize the solutions.

 

[AKG]

see the picture. EDIT: There's a minor mistake in the picture - the labels A and B should switch places.

 

[nothing123]

someone help please.

 

[nothing123]

thanks AKG for the diagram, totally forgot that the theorem applies to angles and REFLEX central angles on the same arc. thank you other posters for your help as well.

i'm having some trouble understanding the word "subtend", especially since it's a major component of circles and circle theorems. for example, the Cyclic Quadrilateral Property states a quadrilaterial is cyclic if and only if one side subtends equal angles at the remaning vertices. what does it mean if a chord subtends an angle(s)?

 

[AKG]

The chord AB subtends angles of the form <APB

 

[nothing123]

i have some more questions if you guys don't mind.

1. BDC is a minor arc of a circle, centre O. CD is produced to E. If ∠BOC = 110°, find the number of degrees in ∠BDE.

basically i need help with setting this up. if BDC is a minor arc, is BC the chord? when it says CD is produced to E, where is E (can it be anywhere in the circle, outside the circle)?

2. A piece of a chariot wheel was uncovered by an archeologist. Describe how you could locate the centre of the original wheel.

http://img83.imageshack.us/img83/8079/chariot2et.jpg

i have no clue, any hints on what circle theorems i can use?

3. A cylindrical water tank, lying horizontally, has an interior length of 10 units and an interior diameter of 6 units. If the rectangular surface of the water has an area of 40 units, what is the depth of the water in the tank?

http://img83.imageshack.us/img83/5999/tank1nb.jpg

is the interior length described the height? do i have to use volume formulas in this or can i simply resort to circle properties?

 

[AKG]

For number 1, the wording confuses me.

For number 2, what can you say about the bisector of a chord? What happens when you look at two chord bisectors?

For number three, you just need circle properties. Why would interior length describe the height? Interior length describes the length! It gives you the interior diamater, so obviously, the length is given in the direction perpendicular to all diameters (i.e. from left to right)

 

[Emieno]

Nothing123, I truly don't understand three of your previous post's problems. But the original ones can be more simple to me and below reasonings may be referenced if you already solved the problem in some other possible ways...

1. A, B and C are points on the circumference of a circle, centre O, and ∠BAC=115°. Calculate the number of degrees in ∠OBC.

You may need to connect O and A to divide O and A into O1, O2 A1, A2. You then can calculate angle O using feature of a triangle (i.e O1+A1+B=180) and since OA=OB=OC which means A1=B and A2=C, you have O1+2A1=180. You do the same with the remaining triangle to have O2+2A2=180. Because A1+A2=115 you then thus have O=130.
You can find then OBC=25

2. ABCD is a quadrilateral inscribed in a circle and AB = CD. Prove that AC = BD.
This is pretty simple I think, if AB = CD, then BC = DA (do I need to say why?).

Your answer to 'why' is your if...I think too it is a truth and needless to prove provided also that you don't forget to include "since given A, B, C , D are on the same circle" in your answer sheet to make it clearer.

 

[Ba]

For number one, assuming the wheel is circular still after years of sitting around I would just use that radii are tangent to the surface of the circle.

 

[nothing123]

thanks guys but the first two initial questions have been already solved. I'm really looking help for the second set questions. AKG, i assumed the interior length was the height of the cylinder if it stood upright; and by your definition, it would be the height.

 

[nothing123]

ok so,

for number 2, chord bisectors are perpendicular to the chord and run through the centre of the circle. we now know any point on the chord bisector can be the centre. by using a second chord bisector, the interesection between two actually yields the exact coordinates of the centre. AKG, is this what you were hinting at?

my solution for number 3 is the following: if the area of the rectangle is 40 and the interior length is 10, the widths (or chords of the circle) are 4 units long. by applying the chord right bisector property, we can say that the the radius of the circle and half the chord form a right angle. therefore, the distance from the chord to the centre is sqrt(3^2-2^2) = sqrt(5). now adding the other half of the diameter, we have found the depth to be 3 + sqrt (5). is this correct?

 

[AKG]

ok so,

for number 2, chord bisectors are perpendicular to the chord and run through the centre of the circle. we now know any point on the chord bisector can be the centre. by using a second chord bisector, the interesection between two actually yields the exact coordinates of the centre. AKG, is this what you were hinting at?

Exactly.

my solution for number 3 is the following: if the area of the rectangle is 40 and the interior length is 10, the widths (or chords of the circle) are 4 units long. by applying the chord right bisector property, we can say that the the radius of the circle and half the chord form a right angle. therefore, the distance from the chord to the centre is sqrt(3^2-2^2) = sqrt(5). now adding the other half of the diameter, we have found the depth to be 3 + sqrt (5). is this correct?

Well done.

 

[civil_dude]

That works out. It also is easy if you use trig, with the hypotenous (radius) = 3, then cos(x) * 3 = 2, therefore depth = 3 + 3 * sin(x) = 5. {x = 48.2 degrees}