circle geometry questions (1 Viewer)

[.ben]

hi i have a few qns:

1. AB and AC are equal chords of a circle. AD and BE are parallel chords through Aand B repectively. Prove that AE is parallel to CD.

2. A, B, C, D are four points in order on a circle, centre O. If angle BAC=28º and angle OBA=53º, calculate the angles OAC, ACB and ADC.

3. Two perpendicular chords AB and CD of a circle, centre O, intersect at E. Prove that angle AOD + angle BOC=180º.

Thanks.

btw they're from Fitzpatrick.

 

[Jago]

i dont suppose these questions had diagrams?

 

[.ben]

i dont suppose these questions had diagrams?

unfortunately no

 

[noah]

first one:

BEA = ADC (subtented by equal chords AB and AC respectively)

BEA = EAD (alternate angles on parallel lines BE and AD)

therefore:
ADC = EAD (both equal BEA)

ADC and EAD are therefore equal and alternate angles of the lines AE and CD

there fore Ae is parallel to CD

 

[.ben]

first one:

BEA = ADC (subtented by equal chords AB and AC respectively)

BEA = EAD (alternate angles on parallel lines BE and AD)

therefore:
ADC = EAD (both equal BEA)

ADC and EAD are therefore equal and alternate angles of the lines AE and CD

there fore Ae is parallel to CD

o thanks noah i didn't think of using the equal chords theorem.

 

[noah]

i wont actually do 2 and 3 i dont have time but for 2:

trynd draw a fairy accurate diagram so that you can see wich side of the centre cirtain chords lie and which angles are acute/obtuse etc

use isosoles triangles formed by two radii and a chod and angle sums of trangles and also the fact that an angle at the centre is twice that at the circumfrence when subtended by the same chord or arc.

 

[Trev]

Q3.
You should be able to figure it out from the diagram and write the proofs and whatnot.
Just use the angle from same arc makes angle twice that at centre than at circumference and sum of angles of a triangle.

 

[.ben]

k thanks noah and trev.

I kept doin questions and found these two more probs:

4. ABCD is a cyclic quadrilateral in which AD=DC=CB. Prove that
a) angle ADC = angle DCB
b) DC || AB

5. C is any point on a circle diameter AB. P and Q are points on the minor arcs AC and BC. Prove that angle APC + angle CQB = 3 right angles.

thanks.

 

[noah]

4a)

let CBA = x

therefore ADC = 180 - x (cyclic quad, supp opp angles)

ADC is isosolese

therefore DAC = ACD = x/2

DBC = DAC = x/2 (angle subyended byh the same seg.)

BCD is isosolese

therefore DBC = BDC = x/2

therefore triangles BCD and ADC are congruent (AAS)

therefore BCD = ADC (corresponding andgle in cong. triangles.)


b) BCD = 180 - DAB (opp angles in cyclic quads are supp.)

therefore ACD = 180 - DAB (as ACD = BCD)

therefore ACD is supp to DAB

therefore DC is parallel to AB (if cointerior angles are supp then the lines are parallel)




5 consider the triangle ACB,

ACB is an right angle (subtended by the diameter AB)

let BAC = x

therefore ABC = 90 - x (angle sum of a tri)

consider the cyclic quad APCB

APC = 180 - ABC = 90 + x (opp angles in cyclic quads are supp)

similarly in cyclic quad ACQB'

CQB = 180 - ACB = 180 - x

therefore APC + CQB = 90 + x + 180 - x = 270 = 3 right angles.

 

[.ben]

hey noah isn't it

'Equal angles at the CENTRE of a circle stand on equal chords'

edit: thanks for the other solutions

 

[noah]

yes but equal angles at the circumfrenced also subtend equal chords.

this is important it will come up a lot