,.,good day,.,amm,.,.how should i draw a quadrilateral which is circumscribable and at the same time inscribable??can u give me examples with measures???thnx a lot
The condition for the quadrilateral to be circumscribable (by a circle) is that both pairs of opposite angles should add up to 180°. The condition for it to be inscribable is that the sum of the lengths of both pairs of opposite sides should be the same. Put those two conditions together and I guess you have the condition given by earboth.
In fact, you can choose any four lengths a, b, c, d satisfying a+c = b+d, and there will be a quadrilateral having those four lengths for its sides which is both inscribable and circumscribable. If the vertices are A, B, C, D, with AB = a, BC = b, CD = c and DA = d, then the length x of the diagonal AC is given by
$\displaystyle \boxed{x^2 = \frac{(ac+bd)(ad+bc)}{ab+cd}}.$
Once you know x, it is easy to draw the quadrilateral. For example, if a=3, b=4, c=6, d=5, then $\displaystyle x\approx5.94$ and the quadrilateral looks like the picture below. (The incircle is not exactly located, because I was guessing the position of its centre and didn't get it quite right.)