F.5 Maths Circle

Draw a st. line PHK//AB, and PHK cuts OM at H and QB at K.

In quad. PAMH :
PA//HM (Given)
AM//PH (drawing)
Hence, PAMH is a //gram. (2 pairs of opp. sides //)
∠PAM = rt.∠ (given)
Hence, PAMH is a rectangle (//gram with a rt.∠ int. ∠)

∠HMB = ∠PAM = rt.∠ (corr. ∠s, PA//HM)

In quad. HMBK :
Since ∠HMB + ∠KBM = 2rt.∠, then HM//KB (consecutive int. ∠s supp.)
MB//HK (drawing)
Hence, HMBK is a //gram (2 pairs of opp. sides //)
∠KBM = rt.∠ (given)
Hence, HMBK is a rectangle (//gram with a rt.∠ int. ∠)

In ΔPQK :
OH//QK (proven)
Hence, PH = HK (intercept theorem)

But PH = AM and HK = MB (opp. sides of rect.)
Hence, AM = MB (axiom)
Since M is the mid-pt. of CD, then CM = MD (given)

AM ‒ CM = MB ‒ MD (axiom)
Hence, AC = DB

"Using centre O as another circle's centre which passes through A and B" is ok.
But first of all, you need to prove why this circle can pass through these 2 points.

2015-06-12 09:15:57 補充：
The simplest way is using "intercept theorem" to prove OM丄AB,
then using "line through centre perpendicular chord bisect chord",
so CM = MD, and then AC = BD.

sorry for forgetting the question does not mention that AB//PQ
I could see that AB is not // PQ in the figure
just try to find if there are any other ways, haha

To YTC,

You are right, the question doesn't mention AB//PQ, so the proof done by Toi Lam is incorrect.