JavaScript seems to be disabled in your browser. For the best experience on our site, be sure to turn on Javascript in your browser. All our lives, we've been taught that boxes are bad. We shouldn't put things in boxes at least, according to Rosencrantz and Guildenstern. It's better to think outside the box.
And it's not exactly a compliment when your grandma tells you that your figure's getting "boxy. A cyclic quadrilateral is a quadrilateral whose vertices all lie on a circle. It doesn't matter if the angles or sides are parallel or congruent or none of the above.
If we can draw a circle that connects all the vertices of that quadrilateral, it's cyclic. No questions asked. The easiest way to figure out which quadrilaterals are cyclic is to simply draw the smallest possible circle around them. When we draw the smallest circles that we can around each quadrilateral, we see that B and C have all of their vertices on the circle.
The quadrilateral in A has only two vertices on its circle. So B and C are cyclic, and A is not. Simple as that. Well, what's so special about cyclic quadrilaterals anyway?
We could stick quadrilaterals into circles all day long, but what's the point? Opposite angles of cyclic quadrilaterals are always supplementary. How's that for a point? And we're just getting started. It's also true that if two opposite sides of a cyclic quadrilateral are congruent, then the other two sides are parallel.
We might not have expected this, but it's true nonetheless. Just like we might not have expected Napoleon Bonaparte to have had ailurophobia a fear of cats.
But he did. Note they each subtend the chord PQ. Since there are congruent circles the arcs the angles subtend have the same measure. You are almost done. Two circles which share a common center are called concentric circles. If you have proved the theorem embedded in part a of Problem 2. Otherwise, there are several ways to consider congruent triangles involving radii of the large circle to A, B, C, and D as well as the perpendiculars from O to the points of tangency with the small circle. A perpendicular to a chord from the center of a circle goes to the midpoint of the chord.
Measurement of the angle COD or the perimeter of triangle PCD will confirm that what you want to prove appears to be the case. Yet, no amount of measuring will establish the proof. Check out if AOBP is a cyclic quadrilateral.
The diagonal AC contains O. OB and OE will surely be helpful segments to use. BAEO is a square with side equal to the radius of the circle. The side DC intersects the circle at Q.
Prove that. Justify your answers. Prove: If a circumscribable isosceles has parallel bases a and b and congruent sides c show that there is a rhombus with sides of length c that is circumscribable about the same circle. A circle O is inscribed in a rhombus. ABCD is a quadrilateral whose vertices are the four points of tangency. What kind of quadrilateral does ABCD seem to be? Open the GSP file and explore.
Or construct a GSP file of your own. It looks, like a rectangle, seems to always be a rectangle as the rhombus is changed, proof? Consider the diagonals of the rhombus and the diagonals of the inscribed quadrilateral. Also, see problem 2. If a rectangle is inscribed in a circle, the circumscribable quadrilateral with points of tangency at the vertices of the rectangle is a rhombus.
Prove the following. Open the GSP file for exploration and hints. The trapezoid is not necessarily isosceles. See GSP file. There are several ways to do this. Then consider the vertical angles at P and argue that center O2 is also on the angle bisector. Prove that the common point of the two tangent circles and the centers are collinear.
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