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?
Please help, thank you!!
RE Toi Lam:
Prove of rectangle:
The question does not mention that AB//PQ.
And also how can you determine ABQP is a rectangle only from PA//OM//BQ?
- 土扁Lv 75 年前最愛解答
Draw a st. line PHK//AB, and PHK cuts OM at H and QB at K.
In quad. PAMH :
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.)
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 :
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
- 少年時Lv 75 年前
"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.
- 5 年前
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
- 邊位都好Lv 55 年前
You are right, the question doesn't mention AB//PQ, so the proof done by Toi Lam is incorrect.