# F.5 Maths Circle

In the figure, PQ is the diameter of the semi-circle with centre O. The straight line AB cuts the semi-circle at C and D. PA⊥AB, QB⊥AB and M is the mid-pt of CD.

Given that PA//OM, prove that AC=BD.

Is the method that " using centre O as another circle's centre which passes

through A and B" wrong?

If it is wrong, then how can I do this question?

RE Toi Lam:

Prove of rectangle:

http://www.regentsprep.org/regents/math/geometry/g...

The question does not mention that AB//PQ.

And also how can you determine ABQP is a rectangle only from PA//OM//BQ?

### 4 個解答

• 最愛解答

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

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)

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.