YTC 發問於 科學及數學數學 · 5 年 前

F.5 Maths Circle

Picture: http://postimg.org/image/ih8mxq0zr/

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!!

更新:

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

評分
  • 土扁
    Lv 7
    5 年 前
    最佳解答

    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

    • 登入以回覆解答
  • 5 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.

    • 登入以回覆解答
  • 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

    • 登入以回覆解答
  • 5 年 前

    To YTC,

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

    • 登入以回覆解答
還有問題嗎?立即提問即可得到解答。