• Mar 20th 2010, 04:16 PM
logan6
Prove that quadrilateral ABCD is cyclic if diagonal AC is perpendicular to side BC and diagonal BD is perpendicular to side AD.

Need a hint to start...I'm stuck with this...

• Mar 20th 2010, 05:38 PM
Soroban
Hello, logan6!

Quote:

Prove that quadrilateral ABCD is cyclic
if diagonal $AC \perp BC$ and diagonal $BD \perp AD.$

Code:

D    *    *    C
o- - - - - - -o
* /  *      *  \ *
* /        *        \ *
/    *      *    \
*/  *              *  \*
A o - - - - - - - - - - - o B

$AC \perp BC$
Then right triangle $ACB$ is inscribed in semicircle $ADCB.$

$BD \perp AD$
Then right triangle $ADB$ is inscribed in semicircle $ADC B.$

Hence, quadrilateral $ABCD$ is inscribed in semicircle $ADCB.$

Therefore: .quadrilateral $ABCD$ is cyclic.

• Mar 21st 2010, 02:01 AM
logan6
Thank you very much! You made my bulb glow!

Note: I made a petite modification. I hope you don't mind.

http://www.mathhelpforum.com/math-he...692b9e4e-1.gif
Then right triangle http://www.mathhelpforum.com/math-he...5f587e85-1.gif is inscribed in circle ACB. (r=AB/2, center in the middle of hypotenuse AB)

http://www.mathhelpforum.com/math-he...01a1a70f-1.gif
Then right triangle http://www.mathhelpforum.com/math-he...02abc75b-1.gif is inscribed in circle ADB.
(r=AB/2, center in the middle of hypotenuse AB))

Hence, quadrilateral http://www.mathhelpforum.com/math-he...84872ca7-1.gif is inscribed in circle http://www.mathhelpforum.com/math-he...ef48d32e-1.gif