Circle Geometry (1 Viewer)

taggs-sasuke · May 22, 2009

I'll try posting here but if I don't get any replies, I'll post a new thread.

Well, I have more Circle Geometry questions...

I'll ask one question at a time so it won't turn you guys off.

Two circles are drawn with centres O and H. A line ACDB is drawn perpendicular to OH, cutting the circles at A, B and C, D respectively (i.e. it is not the common chord). Prove AC = BD.

Thanks guys!

kurt.physics · May 22, 2009

taggs-sasuke said:
I'll try posting here but if I don't get any replies, I'll post a new thread.

Well, I have more Circle Geometry questions...

I'll ask one question at a time so it won't turn you guys off.

Two circles are drawn with centres O and H. A line ACDB is drawn perpendicular to OH, cutting the circles at A, B and C, D respectively (i.e. it is not the common chord). Prove AC = BD.

Thanks guys!

(This is the diagram)

$$Name their intersection Z$

$\angle AZO = \angle BZO = 90^{\circ}\,\,$(ACDB \perp OH)$$

$AO = BO\,\,$(equal radii in circle O)$$

$OZ\,\,$is common$$

$\therefore\,\,$By RHS$,$

$\triangle AZO \cong \triangle BZO$

$\therefore AZ = BZ\,\,$(corresponding sides in congruent triangles)$$

$In circle H:$

$CZ = DZ\,\,$(A perpendicular line from the centre of a circle to a chord bisects the chord)$$

$Combining these two facts:$

$$As$\,\,CZ = DZ$

$$And$\,\,AZ = BZ$

$$Then$\,\, CZ - AZ = DZ - BZ$

$$But$\,\, CZ - AZ = AC$

$$And\,\, DZ - BZ = BD$

$\therefore AC = BD\,\,$as required$$

kurt.physics · May 22, 2009

taggs-sasuke said:
Well, I have more Circle Geometry questions...

More please

taggs-sasuke · May 23, 2009

kurt.physics said:
(This is the diagram)

$$Name their intersection Z$

$\angle AZO = \angle BZO = 90^{\circ}\,\,$(ACDB \perp OH)$$

$AO = BO\,\,$(equal radii in circle O)$$

$OZ\,\,$is common$$

$\therefore\,\,$By RHS$,$

$\triangle AZO \cong \triangle BZO$

$\therefore AZ = BZ\,\,$(corresponding sides in congruent triangles)$$

$In circle H:$

$CZ = DZ\,\,$(A perpendicular line from the centre of a circle to a chord bisects the chord)$$

$Combining these two facts:$

$$As$\,\,CZ = DZ$

$$And$\,\,AZ = BZ$

$$Then$\,\, CZ - AZ = DZ - BZ$

$$But$\,\, CZ - AZ = AC$

$$And\,\, DZ - BZ = BD$

$\therefore AC = BD\,\,$as required$$

Thanks for coming to my rescue!

And yay I get it.

taggs-sasuke · May 23, 2009

kurt.physics said:
More please

Hahaha.

Next question:

Two equal circles intersect in P and Q. A chord APB is drawn through P parallel to the line joining the centres O and H and meeting the circles in A and B. From O and H perpendiculars OX and HY are drawn to AB.

a. Prove OH = XY. What is true about OH and AB?

b. Prove OX = HY and also that the chords AP and PB are equal.

Thanks!

taggs-sasuke · May 23, 2009

I'm going to go against my word but that's only because I really need help with Circle Geometry...

Anyway:

AB and DC are chords of a circle which meet when produced at X. The centre of the circle is O. If angleAOB = 110, angleXBC = 68 and angleXCB = 80, find angleCOD.

Thanks!

taggs-sasuke · May 23, 2009

Another question:

untitled.bmp

Thanks!

kurt.physics · May 23, 2009

taggs-sasuke said:
Hahaha.

Next question:

Two equal circles intersect in P and Q. A chord APB is drawn through P parallel to the line joining the centres O and H and meeting the circles in A and B. From O and H perpendiculars OX and HY are drawn to AB.

a. Prove OH = XY. What is true about OH and AB?

b. Prove OX = HY and also that the chords AP and PB are equal.

Thanks!

$\angle OXP = \angle HYP = 90^{\circ}\,\,$(OX$\,\,\perp\,\,$AB and\,\,$HY$\perp\,\,$AB)$$

$\therefore XO \parallel YH\,\,$(co-interior angles are supplementary)$$

$Furthermore,$

$\angle XOH = \angle YHO = 90^{\circ}\,\,(XY \parallel OH,\,\,$co-interior angles are supp.)$$

$As all angles are right angled and all sides are parallel, then$\,\,XOHY\,\,$is either a square or a rectangle.$

$In either case,$

$OH = XY\,\,$(opposite sides of rectangle/square are equal)$$

$Furthermore,$

$AX = PY\,\,$(A perp. line from the centre to a chord bisects the chord)$$

$Similarly,$

$PY = BY$

$$As$\,\,GH = XY,$

$$then$\,\, PX + PY = OH$

$$But$\,\,AP = 2PX\,\,$and$\,\,BP = 2PY$

$\therefore 2\cdot OH = 2(PX + PY)$

$\therefore 2 OH = AB$

$$(Since$\,\,2(PX + PY) = AB)$

$In other words, AB is twice the size of OH$

$We have shown that XOHY is rectangular,$

$\therefore OX = HY\,\,$(opp. sides of rectangle are equal)$$

kurt.physics · May 23, 2009

taggs-sasuke said:
I'm going to go against my word but that's only because I really need help with Circle Geometry...

Anyway:

AB and DC are chords of a circle which meet when produced at X. The centre of the circle is O. If angleAOB = 110, angleXBC = 68 and angleXCB = 80, find angleCOD.

Thanks!

$First we have to understand what the diagram looks like, the key word to understand this is 'produced at X'. This means that the two chords do not intersect inside the circle, but outside! So now you just draw the diagram with this information.$

$Construct the lines AO, BO, DO, CO and BD$

$Note$\,\, AO = BO = CO = DO\,\,$(equal radii)$

$\angle BAO = \angle ABO = 35^{\circ}\,\,(AO=BO,\,\,$base angles issos$ \,\,\triangle\,)$

$\therefore \angle DBO = 77^{\circ}\,\,$(angles on a straight line)$$

$\therefore \angle BDO = 77^{\circ}\,\,$(base angle of issos.$\,\, \triangle)$

$\therefore \angle CDO = 23^{\circ}\,\,$(angles on a straight line)$$

$\therefore \angle DCO = 23^{\circ}\,\,$(base angle of issos.$\,\, \triangle)$

$\therefore \angle COD = 134^{\circ}\,\,$(angle sum$\,\, \triangle)$

kurt.physics · May 23, 2009

taggs-sasuke said:
Another question:

untitled.bmp

Thanks!

$This question works all on the fact that the angle at the centre of a circle is twice the angle at the circumference and also that the angle sum of a quad is 360 degrees. Can you figure it out now?$

taggs-sasuke · May 24, 2009

kurt.physics said:
$First we have to understand what the diagram looks like, the key word to understand this is 'produced at X'. This means that the two chords do not intersect inside the circle, but outside! So now you just draw the diagram with this information.$

$Construct the lines AO, BO, DO, CO and BD$

$Note$\,\, AO = BO = CO = DO\,\,$(equal radii)$

$\angle BAO = \angle ABO = 35^{\circ}\,\,(AO=BO,\,\,$base angles issos$ \,\,\triangle\,)$

$\therefore \angle DBO = 77^{\circ}\,\,$(angles on a straight line)$$

$\therefore \angle BDO = 77^{\circ}\,\,$(base angle of issos.$\,\, \triangle)$

$\therefore \angle CDO = 23^{\circ}\,\,$(angles on a straight line)$$

$\therefore \angle DCO = 23^{\circ}\,\,$(base angle of issos.$\,\, \triangle)$

$\therefore \angle COD = 134^{\circ}\,\,$(angle sum$\,\, \triangle)$

I overlooked the isosceles triangle AOB.

Thanks!

taggs-sasuke · May 24, 2009

$In other words, AB is twice the size of OH$

$We have shown that XOHY is rectangular,$

$\therefore OX = HY\,\,$(opp. sides of rectangle are equal)$$

I understand why AB is twice the size of OH but how does that make XOHY a rectangle?

kurt.physics · May 24, 2009

taggs-sasuke said:
$In other words, AB is twice the size of OH$

$We have shown that XOHY is rectangular,$

$\therefore OX = HY\,\,$(opp. sides of rectangle are equal)$$

I understand why AB is twice the size of OH but how does that make XOHY a rectangle?

It doesnt make XOHY a rectangle, but i had showed that XOHY was rectangular here:

kurt.physics said:
$As all angles are right angled and all sides are parallel, then$\,\,XOHY\,\,$is either a square or a rectangle.$

The only reason why i wrote

$We have shown that XOHY is rectangular,$

was because i wanted to reiterate what i said as a reason for the next line

$\therefore OX = HY\,\,$(opp. sides of rectangle are equal)$$

But i could/should have said

$We have shown that XOHY is either a rectangle or a square, in either case:$

taggs-sasuke · May 24, 2009

I'm sorry...

I get it now.

Thanks.

taggs-sasuke · May 24, 2009

Can I have another clue?

I've found more numbers. (

)

kurt.physics · May 24, 2009

taggs-sasuke said:
Can I have another clue?

I've found more numbers. ( )

For angle x, you must use the fact that the angle at the centre of a circle is twice the angle at the circumference.

So the angle opposite x is 140, so the angle at the centre is 360 - 140 = 220. So 2x = 220, so x = 110

The angle sum of a quadrilateral is 360, so

y + 140 + 110 + 62 = 360

y + 312 = 360

y = 48

taggs-sasuke · May 24, 2009

kurt.physics said:
For angle x, you must use the fact that the angle at the centre of a circle is twice the angle at the circumference.

So the angle opposite x is 140, so the angle at the centre is 360 - 140 = 220. So 2x = 220, so x = 110

The angle sum of a quadrilateral is 360, so

y + 140 + 110 + 62 = 360

y + 312 = 360

y = 48

Oh...

Thanks.

taggs-sasuke · May 24, 2009

Quick question:

x = black dot
y = red dot

... but what reason/theorem is it?

(Is it 'opposite angles of a cyclic quadrilateral are supplementary'? But is it a cyclic quadrilateral?)

Thanks!

kurt.physics · May 24, 2009

taggs-sasuke said:
Quick question:

x = black dot
y = red dot

... but what reason/theorem is it?

(Is it 'opposite angles of a cyclic quadrilateral are supplementary'? But is it a cyclic quadrilateral?)

Thanks!

Is there any more information?

taggs-sasuke · May 24, 2009

kurt.physics said:
Is there any more information?

No.

But this exercise comes under cyclic quadrilaterals and concyclic points.

Edit:

Don't worry. I have the solution.

Circle Geometry (1 Viewer)

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