1. ## Analytic proof

Can I already assume that the quadrilateral that is being referred here is a parallelogram?
"A quadrilateral whose diagonals bisect each other at right angles is a rhombus."

Please take note that we were asked to prove the above statement.

I don't even know how to start my proof.

2. Originally Posted by mamen
Can I already assume that the quadrilateral that is being referred here is a parallelogram?
"A quadrilateral whose diagonals bisect each other at right angles is a rhombus."

Please take note that we were asked to prove the above statement.

I don't even know how to start my proof.

I forgot to label the point at the center. Call it E.

First, can you show that triangles AEB and CED are congruent? This means that AB = CD.

Next, a little tougher. Can you show that triangles AED and AEB are congruent? This means that AB = AD.

Similar to the first problem, show that triangles AED and CEB are congruent. Thus AD = CB.

Putting it all together gives AB = BC = CD = DA. So we have a quadrilateral such that all sides are equal. What kind of quadrilateral is this?

-Dan

I can't get the image to upload. Here's a quick rundown. I've got a quadrilateral ABCD. I have diagonals AC and BD meeting at right angles at a point E. The diagonals are bisected so we know that BE = ED and AE = EC.

3. Sir, I know how to prove this: "A quadrilateral whose diagonals bisect each other at right angles is a rhombus." by using column method but I really don't have the idea on how to prove this analytically. I really don't know how to start my proof, and I can't even think of the proper coordinates to be used. thanks in advance, sir...

4. Originally Posted by mamen
Sir, I know how to prove this: "A quadrilateral whose diagonals bisect each other at right angles is a rhombus." by using column method but I really don't have the idea on how to prove this analytically. I really don't know how to start my proof, and I can't even think of the proper coordinates to be used. thanks in advance, sir...
All right then. Again we have a quadrilateral ABCD. Again the point of intersection of the two diagonals is E. I am going to call AE = EC = R and BE = ED = r. Notice that AEB is a right triangle, so the side AB (the hypotenuse) is sqrt(R^2 + r^2). Notice, in fact, that all sides can be calculated this way and AB = BC = CD = DA = sqrt(R^2 + r^2). The result follows.

-Dan

Edit: If you insist on using a coordinate system, then you have to find coordinates for A, B, C, and D then calculate AB, BC, CD, and DA. Believe me, you will want to see if you can use the right triangle approach above. Using coordinates is a terribly complicated way to do this problem and stretches on for about a page of work.

5. A geometric proof is easy. There are two perpendicular bisectors each bisecting the other.All sides are then equal. Try it.An analytic solution is involved. Draw a line segment on a grid.Locate its midpoint and slope.Erect the perpendicular of this segment by drawing a line at the midpoint with inverse slope of the segment.From the midpoint mark two equidistant points above and below the segment.Using the distance formula show that the four points are equal

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# Using analytical geometry prove that the diagonal of a rhombus are perpendicular to each other

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