Yahoo 知識+ 將於 2021 年 5 月 4 日 (美國東岸時間) 停止服務，而 Yahoo 知識+ 網站現已轉為僅限瀏覽模式。其他 Yahoo 資產或服務，或你的 Yahoo 帳戶將不會有任何變更。你可以在此服務中心網頁進一步了解 Yahoo 知識+ 停止服務的事宜，以及了解如何下載你的資料。

Fireman
Lv 7
Fireman 發問於 Science & MathematicsMathematics · 1 十年前

ABC is a triangle with angle C =90*.CD is drawn &D lies on AB such that angleCDA=angleCDB=90*.?

Prove that AC^2/BC^2=AD/BD

更新:

no proper reply yet. where are the top contributers?

4 個解答

相關度
  • Sridhar R
    Lv 6
    1 十年前
    最愛解答

    I think you have asked this question twice i have already answered this anyhow here is the proof

    in triangle ACD

    angle ACD + angle CAD = 90

    in triangle ACB

    angle CAB + ABC = 90

    angle CAB = angle CAD same angle

    therefore angle ACD = angle ABC --------------------------------(1)

    consider the Triangles

    ACD and CBD

    angle ACD = angle CBD | angle ABC = angle CBD

    angle ADC = angle CDB

    so the two triangles are similar

    so we have

    AD/CD = AC/CB = DC/DB -----------------------------------2)

    AD/DC = AC/BC

    DC/BD = AC/BC | from 2

    multiplying both

    AD/DC * DC/BD = AC/BC * AC/BC

    AD /BD = AC^2/BC^2 prooved

  • Gufran S
    1 十年前

    we know that in triangle ABC AC^2/BC^2=DC^2+AD^2/CD^2+BD^2

    =>YOUR ANSWER

    資料來源： general
  • 匿名
    1 十年前

    Math is dull lol

  • saaranshbhalla
    1 十年前

    right

還有問題嗎？立即提問即可得到解答。