In quadrilateral abcd, AC=ad & bc=cd. The shortest distance of AC from d is dp. Prove b, p, d is collinear?
In quadrilateral abcd, AC=ad & bc=cd. The shortest distance of AC from d is dp. Prove b, p, d is collinear?
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Please show your work, so that we can see what you've already tried.
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"English is the most ambiguous language in the world." ~ Yours Truly, 1969
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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