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Initially I was of the opinion that if two diagonals in a parallelogram intersect then the angle formed at the point of intersection is 90 degrees (I came up with this conclusion by inserting values) , if this applied to a parallelogram then I assumed it also applied to a square or a rectangle.In both the figures below i assumed angles A , B , C will always be 90 degrees at least that's what I thought.Am i wrong ?

enter image description here

Now I just came across a question which is conflicting with this concept the question is:

ABCD is rectangle the diagonals AC and BD intersect at E (reflected in red figure) .Which of the statement is not necessarily true. The answer to this was : AE is perpendicular to BD. Could anyone let me know why is this not true ? Did i have the wrong idea to start with ?

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Draw (with a ruler) a long skinny rectangle, say base $6$, height $1$. Now draw the diagonals. You will see that the two diagonals definitely do not meet at right angles, not even close!

It turns out that a parallelogram has its diagonals meeting at right angles if and only if the parallelogram is a rhombus (all sides equal). Note that a square is a special case of a rhombus.

Proof: It is fairly easy to prove that the diagonals of a parallelogram (and therefore of the special parallelogram called a rectangle) bisect each other.

Let the two diagonals of a parallelogram have length $2p$ and $2q$ respectively. If the angle at which they meet is $90^\circ$, then by the Pythagorean Theorem each side of the rectangle has length $\sqrt{p^2+q^2}$. So in particular all the sides of the parallelogram are equal, that is, we have a rhombus.

Conversely, by a congruent triangles argument, if we have a rhombus, its diagonals meet at right angles.

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  • $\begingroup$ What about parallelograms ? Square ? $\endgroup$ – MistyD Jun 19 '12 at 3:20
  • $\begingroup$ " It turns out that a parallelogram has its diagonals meeting at right angles if and only if the parallelogram is a rhombus (all sides equal)" - Thanks for clearing this up $\endgroup$ – MistyD Jun 19 '12 at 3:26
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To clear it out , take example of a 30 cm ruler scale. It is rectangle in shape and definitely a parallelogram(all rectangles are parallelogram).

Now make diagonals. You will find that the angles made at the diagonals intersections are definitely not 90 (intersecting angles - opp angles are same . Out of which one set's value is way too high than 90 and others very much less than 90 )

Now slowly imagine that the breadth of ruler is increasing and length is decreasing such that at a point all sides become equal and the new shape formed is square (which is also a rhombus)

In that square you will find that the angles formed by diagonals intersection are all same and 90 )

After this if you keep on increasing value of breadth and reducing value of breadth then these intersection angles again start varying (actually the values will start swapping with values prior to reaching to square shape).

After this visualization you can conclude that the intersection angles formed by diagonals are 90 only if the shape is square w.r.t rectangle or when the shape is rhombus w.r.t parallelogram.

Breaking the myth - Diagonals of all parallelogram do not bisect the vertices angles. They bisect only in case of square or rhombus.

Hope this helps

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