# Thread: Circle Geometry Extension 1 Help

1. ## Circle Geometry Extension 1 Help

1. Let A, B, C, D, and E be five points in order around a circle with centre O, and let AOE be a diameter. Prove that <ABC + <CDE = 270◦.

2a) Prove that if two chords of a circle bisect each other, then they are both diameters.
b) Prove that if the chords AB and PQ intersect at M and MA = MP, then MB = MQ, BP = AQ and AP ∥ QB.

Can someone solve these questions? Thanks

2. ## Re: Circle Geometry Extension 1 Help

Q1

Draw diameter AE. Pick any arbitrary points B, C and D between A and E on, say, the upper semi-circle (I used clockwise order). Join B and E. Then angle EBC + angle CDE = 180 (BCDE being cyclic). Angle ABE = 90 (angle in a semi-circle)

Adding, angle ABC + angle CDE = 270.

3. ## Re: Circle Geometry Extension 1 Help

Q2

a) Let chord AB bisect chord CD (so A, C, B and D are in a clockwise order)

Then we can use following properties to prove AB and CD are diameters:

i) if AB bisects CD, then ACBD is a parallelogram.

ii) since ACBD is cyclic, opp angles are supplementary
.: angle ACB + angle BDA = 180

iii) but ACBD is a //gram so that opp angles are equal

iv) .: angle ACB = angle BDA = 90

v) in cyclic quad ACBD, since angle ACB = 90, then AB must be a diameter (angle in a semi-circle = 90)

vi) .: ACBD must be a rectangle and .: angle CAD = 90 = angle in a semi-circle & .: CD is a diameter

4. ## Re: Circle Geometry Extension 1 Help

Q2

b) it can be easily shown triangle AMP ||| triangle BMQ

Since AM = MP, AMP is isosceles and .: BMQ is also isosceles and .: MB = MQ so that the 4 base angles are all equal.

Since ang APM = ang PQB and they are alternate angles, AP//BQ

Now triangles BAQ & QPB are congruent, noting ang BAQ = ang QPB and AB = PQ and BQ = QB

.: corresponding sides AQ = BP

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•