# Re: Equidistant Midpoint of a Hypoteneuse

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Posted by Joel on November 05, 2002 at 22:21:24:

In Reply to: Equidistant Midpoint of a Hypoteneuse posted by James on November 05, 2002 at 21:22:07:

: I'm enrolled in a online school and my text is online. The text is very vague in showing me how to do stuff. I am having trouble with a question. If anyone could help it would be much obliged. The question is:

: A right-angled triangle has vertices A(3,4), B(7,-4), C(-5,0). Show that the midpoint of the hypotenuse is equidistant from each vertex.

Ok, that sounds like a silly question. Isn't the midpoint of a line segment by definition the point that is equidistant from each end? But, let's humor them. You can draw some lines on the graph & then either use the Pythagorean theorem to calculate the length of each half. Or, you can draw some lines & then just argue the point logically.

Call point (1,0) D. Call point (1,-2) E. Call point (1,-4) F. Draw line FB from (1,-4) to (7,-4). Draw line DF from (1,0) to (1,-4). Line BF (part of the line y = -4) is obviously parallel to the X axis, so angle DCE is equal to angle EBF (alternate interior angles theorem). Also, angle CDE equals angle EFB (same reason, or, they are both right angles since line x=1 is perpendicular to lines y = 0 and y = -4). Line CD (6 units along x axis from (-5,0) to (1,0) equals line FB (6 units along line y=-4 from (1,-4) to (7,-4). So, by angle-side-angle, triangles CDE and BFE are congruent. Therefore, hypotenuse CE must equal hypotenuse EB.

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