Smile 發問於 科學及數學植物學 · 1 年前

Maths problem, how to do thanks?

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  • SC147
    Lv 6
    1 年前
    最愛解答

    Solution (REVISED) :

    ===============

    ∵ AB=AP, ∴△BAP is an isosceles △

    => ∠ABP=∠APB

    Let ∠ABP=∠APB = b

    From △ABQ,

    ∠AQB=90⁰ (∠ in semicircle)

    ∴ ∠BAQ = 90⁰ - b . . . . . . . . . . . . . ①

    From △PQN,

    ∠QNP=90⁰ (Given that QN丄AP)

    ∴ ∠PQN = 90⁰ - b . . . . . . . . . . . . . ②

    ①&② : ∠BAQ = ∠PQN . . . . . . . . . . . . . (#)

    Now, from △ABQ, ∠ABQ = 90⁰ - ∠BAQ. . . . . . . . . . . ③

    Also, from pt. Q, ∠AQN = 90⁰ - ∠PQN. . . . . . . . . . . . .④

    Concluding from (#), ③&④ : ∠ABQ = ∠AQN

    Conclusion :

    QN is tangent to the circle at Q (converse of ∠ in alternate segment)

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