Prove the concyclic(very hard)

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  • 7 年前
    最愛解答

    a)

    ∠NAP =∠NA1P = 90˚ (given)

    So AA1PN are concyclic (converse of ∠s in the same segment)

    b)

    M is the mid-pt of BC and B1C1 (given)

    △ABC≅△A1B1C1 (given)

    BC=B1C1 (corr.sides,≅△s)

    BM+MC=B1M+MC1

    2MC = 2MC1

    MC=MC1

    AC=A1C1 (corr.sides,≅△s)

    ∠C=∠C1 (common ∠)

    So △AMC≅△A1MC1 (S.A.S.)

    ∠MAC=MA1C1 (corr.∠s,≅△s)

    ∠MAP=MA1P

    So AA1PM are concyclic (converse of ∠s in the same segment)

    c)

    ∠APM=∠AA1M (∠s in the same segment)

    ∠APN=∠AA1N (∠s in the same segment)

    ∠APM-∠APN=∠AA1M-∠AA1N

    ∠NPM=∠NA1M

    So A1PMN are concyclic (converse of ∠s in the same segment)

    ∠NMP + ∠NA1P =180˚ (opp. ∠s, cyclic quad.)

    ∠NMP + ∠NAP= 180˚

    So APMN are concyclic (opp.∠s, supp.)

    Understand?

    資料來源: Myself:)
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