ABCD is a cyclic quadrilateral whose two diagonals are perpendicular.

Question

ABCD is a cyclic quadrilateral whose two diagonals are perpendicular. If $R$ is the radius of the circumcircle, prove that:

$AB$^$2$ + $BC$^$2$ + $CD$^$2$ + $DA$^$2$ = $8R$^$2$

If the centre of the circle is O, I tried drawing radii perpendicular to each of the sides. That way, we get 3 pairs of congruent right triangles.

the diagonals become the diameters of the circle, and then use Ptolemy's theorem — person, Nov 22 '20 at 18:43
If the centre of the circle is O, I tried drawing radii perpendicular to each of the sides. That way, we get 3 pairs of congruent right triangles. — , Nov 22 '20 at 18:44
My bad, I meant that the perpendicular bisectors of the diameters — person, Nov 22 '20 at 18:56

score 4 · Accepted Answer · edited Nov 23 '20 at 00:06

4

Lemma. If $ABCD$ is a cyclic orthodiagonal quadrilateral centered at $M$, then $$\angle AMB+\angle CMD=\angle BMC+\angle DMA=180^\circ$$

Here you have a "proof without words"

Thus, we have that $DA$ and $BC$ form a right triangle with a diameter, and hence $DA^2+BC^2=4R^2$. We similarly obtain $AB^2+CD^2=4R^2$...

edited Nov 23 '20 at 00:06

Angelo

answered Nov 22 '20 at 19:52

Dr. Mathva

Very nice solution. May I ask what did you use to generate the image? – jjagmath Nov 22 '20 at 20:13
@jjagmath, I used Geogebra. – Dr. Mathva Nov 22 '20 at 21:14
Very nice proof, congratulations (+1) – Angelo Nov 23 '20 at 00:03

score 2 · Answer 2 · answered Nov 22 '20 at 20:32

2

Here's my solution.

Use this construction

to prove that $AB = 2R\cos(CAD)$.

By the law of sines we have $CD = 2R\sin(CAD)$, so $AB^2 + CD^2 = 4R^2$.

answered Nov 22 '20 at 20:32

jjagmath

2 Answers2