Talking about shapes, one of the most familiar is the parallelogram. And about it we have a problem as follows.
Given quadrilateral ABCD, E, F is the midpoint of AD, resp. DC. Lines BE and AF intersect at point G. (And the figure is as below.)
Suppose it happens that GE : GA : GF : GB = 1 : 2 : 3 : 4. Attempt to prove ABCD is a parallelogram.
How can the division ratio GE : GA : GF : GB as given, be used as the hints to deduct the parallelogram?
Well, to put the known condition in proper perspective is important. If first attention is directed to GA : GF = 2 : 3, you might wonder how that can be utilized. Suppose you notice that GE : GA = GA : GB = 1 : 2 : 3 : 4, and suppose AF and BE are perpendicular to each other, then we conclude that the two triangles EGA and AGB are similar right triangles .. But how that can be used to help recognize a parallelogram?
By observing AF = AG + GF, and BE = BG + GE, and we find that AF = BE (by applying properties of the proportion, AF : BE = (2 + 3) : (4 + 1) = 1 : 1). So we have two hanging line segments AF and BE that are congruent (equal in length), and both segments are drawn from one vertex (of ABCD) to midpoint of a non-adjacent side. The information in its current form seems much helpful.
Now passing through D let’s draw a parallel line of BE; suppose the line intersects BC at point J: DJ // BE. Suppose that DJ intersects AF at point Q (Q should be between G and F). We work out that DQ = 2 EG (notice E is the midpoint of AD), as well as GQ = AG. So if it held that DQ : QJ = 2 : 3, we would prove that DJ = BE.
(Why doing that? If BE = DJ were proved, then JDEB would be a parallelogram. Parallelogram JDEB implies that AD // BC. Can you see that? Just note ED // BJ, and segment ED is on line AD, resp. BJ is on BC; so we have, finally, AD // BC). So we’ve proved that AD and BC, as one pair of the opposite sides of a quadrilateral, is parallel. We can analogously attempt for AB // CD. Now by definition, ABCD is a parallelogram.
However to work out DQ : QJ = 2 : 3 some effort is needed, and the good news is we are on the track.
A rather interesting way (of course challenging but a pleasant experience), isn’t it? For anyone who wants to do more systematically, or automatically, is there another way?
Yes, there is. That way is to use coordinate system.
Suppose the two hanging segments AF and BE are perpendicular to each other, and let their intersection (G) be the origin: so G(0,0). With the given GE : GA = GA : GB = 1 : 2 : 3 : 4 (we’ll call it1-2-3-4 condition), we draw the following diagram that involves G, E, A, B, F to help us understand, and to figure out coordinates to be E(0,1), A (2,0), F (-3,0) and B (0,-4).
Coordinates of the other two vertices can be worked out from the conditions that E is the midpoint of AD, as well as F is the midpoint of DC. (Noting the positive direction of x-axis points from G to A.) We worked out D (-2,2) and C (-4,-2). If it’s indeed that ABCD is one parallelogram, then the two diagonals AC and BD bisect each other, which means that the midpoint of the two diagonals are the same, as verified straightforwardly:
Midpoint of AC: ((2-4)/2, (0-2)/2) ) = (-1, -1)
Midpoint of BD: ((0-2)/2, (-4+2)/2 ) = (-1,-1)
Since AC and BD bisect each other, ABCD is a parallelogram. But what if the AB and BE are not perpendicular? Well ABCD is still a parallelogram in case – but explanation is needed – for which we can pursue later. It deserves to point out, by changing the x- and y- axis from being orthogonal to arbitrary, whether the shape is parallelogram or not will have the same answer.
Let me ask if you would like this problem. If like, then share it with your friends!
Continuing from the problem discussed above, suppose that apart from AE, BF, we draw CJ and DK, where J is midpoint of AB and K is the midpoint of BC. Here is your turn to take a ramp-up challenge:
(If you truly understand what’s going on in the above discussion, then it’s just a piece of cake!)
(i) Can you show that there are exactly four intersections among the four lines AE, BF, CJ and DK; and the four intersection points connect to make a parallelogram?
(ii) What type of parallelogram is the one mentioned in question (i)? Is it a rectangle or rhombus?
*(iii) Is it possible for the original parallelogram to be a rectangle ABCD? In that case, if GE : GA : GF : GB = 1 : 2 : 3 : 4, what would be the aspect ratio AB : BC (aspect ratio is the ratio of width-to-length in a rectangle shape).