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Thread: Geometry:In quadrilateral abcd, AC=ad & bc=cd. The shortest distance of AC from d is

  1. #1

    Geometry:In quadrilateral abcd, AC=ad & bc=cd. The shortest distance of AC from d is

    In quadrilateral abcd, AC=ad & bc=cd. The shortest distance of AC from d is dp. Prove b, p, d is collinear?

  2. #2
    Elite Member mmm4444bot's Avatar
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    How far did you get? What happened?

    Please show your work, so that we can see what you've already tried.

    Also, kindly read the forum guidelines. Thanks!
    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  3. #3
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    Quote Originally Posted by phoenix101 View Post
    In quadrilateral abcd, AC=ad & bc=cd. The shortest distance of AC from d is dp. Prove b, p, d is collinear?
    The usual convention is to use capital letters for points and lower case for lines/lengths. You seem to be ignoring case (and perhaps letting autocorrect "fix" things). As I understand it, you mean this:

    In quadrilateral ABCD, AC=AD & BC=CD. The shortest distance of AC from D (that is the shortest segment from D to AC) is DP. Prove B, P, and D are collinear.

    But when I sketch the figure, I find that the conclusion is not true. Please check that you stated the problem correctly.

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