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Bisecting a diagonal in a cyclic quad

Let ABCD be a cyclic quadrilateral with side-lengths AB=a, BC=b, CD=c, and DA=d. Let AC and BD be the diagonals. Then AC bisects BD if and only if ab=cd, and BD bisects AC if and only if ad=bc.

Let ABCD be a convex quadrilateral whose diagonals intersect at E. If ABCD is cyclic, PROVE that \frac{AE}{EC}=\frac{ad}{bc} and \frac{BE}{ED}=\frac{ab}{cd}.

Use similar triangles.

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Let ABCD be a convex quadrilateral whose diagonals intersect at E. If \frac{AE}{EC}=\frac{ad}{bc} and \frac{BE}{ED}=\frac{ab}{cd}, PROVE that ABCD is cyclic.

This is the converse of the previous statement in example 1. One proof of it uses similar triangles and the law of sines.

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Let ABCD be a convex cyclic quadrilateral whose diagonals intersect at E. PROVE that diagonal AC bisects diagonal BD if, and only if, ab=cd.

By example 1 we have that \frac{BE}{ED}=\frac{ab}{cd}. If AC bisects BD, then BE=ED. And so ab=cd. Similarly for the converse.

Let ABCD be a convex cyclic quadrilateral whose diagonals intersect at E. PROVE that diagonal BD bisects diagonal AC if, and only if, ad=bc.

By example 1 we have that \frac{AE}{EC}=\frac{ad}{bc}. If BD bisects AC, then AE=EC. And so ad=bc. Similarly for the converse.

Among convex cyclic quadrilaterals, PROVE that only rectangles have the two diagonals bisecting each other.

This follows from example 3 and example 4 above, since ab=cd and ad=bc yield a=c and b=d.

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Takeaway

Let ABCD be a convex cyclic quadrilateral. Then the following statements are equivalent:

  1. ABCD is a rectangle
  2. diagonals AC and BD bisect each other.

Task

  • (Easy verification) Let ABCD be a convex cyclic quadrilateral with vertices at A(-7,7), B(-6,0), C(0,0), D(2,4). Verify that:
    1. diagonal AC bisects diagonal BD
    2. 2AC^2=AB^2+BC^2+CD^2+DA^2
    3. BD^2=\frac{2(AB^2+DA^2)(BC^2+CD^2)}{AB^2+BC^2+CD^2+DA^2}.