Proof:
Given: lines DF and BC are parallel
segments DB and DE are congruent
Show: <FBC trisects <ABC
Let x denote the measure of < FBC ( m<FBC). <FBC and <BFD are congruent because they are alternate interior angles, so m<BFD = x. Segments DE and EF are congruent because both are radii of the circle with center E. Therefore triangle DEF is isosceles, and <EDF is congruent to <EFD (same as <BFD) since both are bases angles. So m<EDF = m<EFD = x. Since the sum of the measures of the angles of a triangle is 180, m<DEF = 180 — 2x. <DEB is supplementary to <DEF, so m<DEB = 180 — (180 — 2x) = 2x. Since it is given that segments DB and DE are congruent, triangle DBE is isosceles. Therefore, <DBE is congruent to <DEB because both are bases angles. So, m<DBE = 2x. The angle < DBE is the same as <ABF, so m<ABF=2x. By angle addition, m<ABC = m<ABF + m<FBC = 2x + x = 3x. So m<ABC is three times m<FBC. In other words, m<FBC is one third of m<ABC. Thus, we have shown that <FBC trisects <ABC.
Discussion:
It is well known that it is impossible to trisect angle as a compass and straightedge construction.
The trisection carried out in our GSP exploration was possible because the software allows dynamic changes that cannot be carried out with the compass and straight edge alone. Manipulating point E to the appropriate place was the crucial aspect. This manipulation is the beyond the reach of standard compass and straight edge constructions because we were looking for the place to put the circle's center so that ray BE, circle E, and line DF all coincided.
Observation:
As P moves the shape of the overlapping area changes. When two sides of square OP line up with the diagonals of square ABCD, it is easy to see that the overlapping area of square OPQR with the yellow square ABCD is one fourth of the total area of the yellow square.
In fact, this ratio holds whenever the squares overlap "cleanly" by intersecting in exactly two points. A proof for this result in the more general case follows. When OP is so short that the overlap is not "clean," OPQR overlaps varying smaller portions of ABCD, causing the ratio
area ABCD / area OPQR to increase.
Proof:
Given: square ABCD
the diagonals of square ABCD intersect at O
square OPQR intersecting square at only two points
Show: area of overlap = one fourth area of yellow square ABCD
Suppose for the purpose of clarity that OP intersects side DC at point S. (A similar argument would hold using each of the other sides: AD, BA, and CB.) Label the intersection of OR with side BC as T. Draw in the segment ST.
Since ABCD is a square, its diagonals are perpendicular. So m<BOC = 90. Let y denote m<BOR. Then its complement m<ROC = 90 - a. But <POR is also a right angle since OPQR is a square. So <POC is complementary to <ROC, and <POC = 90 — (90 — a) = a. Thus <POC (same angle as <SOC) and <BOR (same angle as <BOT) are congruent. Other properties of a square include congruent diagonals that bisect each other and bisect the square's angles. These tell us that segments OC and OB are congruent and that <OCD and <OBC both have measure 45. Therefore triangle SOC and triangle TOB are congruent by ASA. Then by CPCTC, we have segments OS and OT congruent, as well as segments SC and TB.
Let x be the length of a side of square ABCD. Let y = SC = TB. Then CT = CB — TB = x-y.
Then segment ST is the hypotenuse of the right triangle SCT. So ST^2 = SC^2 + CT^2 by the Pythagorean Theorem. Substituting and simplifying, we have ST^ = 2y^2 +x^2 - 2yx.
Let w = OS = OT. ST is also the hypotenuse of triangle SOT. So ST^2 = OS^2 + OT^2 = 2w^2.
Combining the two equations for ST and simplifying, we have w^2 = y^2 + (1/2)x^2 - yx.
Since triangles SOT and SCT are right triangles, their areas are (1/2)(SC)(CT) = (1/2) y(x — y) and (1/2) (OS)(OT) = (1/2) w^2 = (1/2) (y^2 + (1/2) x^2 - yx) , respectively. The sum of the areas of triangles is the area of the overlapping area. So area of overlap = (1/2) y(x — y) + (1/2) (y^2 + (1/2) x^2 - yx) = (1/4) x^2 , which is one fourth the area of square ABCD.