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Session 5: Solutions
 
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Solutions for Session 5, Part C

See solutions for Problems: C1 | C2 | C3 | C4 | C5 | C6 | C7


Problem C1

By construction, E is a midpoint of BC, so the sides CE and EB are congruent. Similarly, the sides DE and EF are congruent. By Fact 1, the angles between the congruent pairs of sides are equal, since they are vertical angles (the two angles at the shared vertex E). Then, by Fact 2, the two triangles are congruent.

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Problem C2

Since the triangles DEC and FEB are congruent (by Problem C1), the corresponding sides in the two triangles are congruent. In particular, DC and FB have the same length. But DC has the same length as AD, since, by construction, D is the midpoint of AC. Therefore, AD and BF have the same length.

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Problem C3

The angles DCE and FBE have the same measure since they are corresponding angles in congruent triangles. If we extend line segments DC and FB into lines, the two angles become alternate interior angles along a transversal to the two lines. Since the alternate interior angles are congruent if and only if the two lines are parallel, we conclude that the two lines and the segments on them are indeed parallel. So we have shown that the segments AD and BF are parallel.

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Problem C4

By Problems C2 and C3, we know that the segments AD and BF have the same length and are parallel. They are also opposite sides in the quadrilateral ABFD. Since we have one pair of opposite sides congruent and parallel, we can use the fourth definition in Fact 3 to conclude that ABFD is a parallelogram.

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Problem C5

Since AB and DF are opposite sides of a parallelogram (by Problem C4), they are parallel. We also know that DE lies on DF since the points D, E, and F all lie on the same line (by construction). Therefore, DE is also parallel to AB.

Finally, since AB and DF are opposite sides of a parallelogram (by problem C4), they must have the same length. Also, DE and EF have the same length by our construction. This means that DE must be half as long as DF (which is DE + EF). Since DE is half as long as DF, it is also half as long as AB.

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Problem C6

Since EF is a midline of ABC, EF is parallel to AC and half the length of AC. Similarly, since HG is a midline of ACD, HG is parallel to AC and half its length. So EF and HG are parallel (since they are both parallel to AC) and of equal length (since they are both half the length of AC). The quadrilateral EFGH is a parallelogram, by Fact 3, since it has a pair of congruent and parallel opposite sides, namely EF and HG.

Thinking even more deeply, you might notice that ABCD was constructed to be a kite, so its diagonals are perpendicular. Therefore, the sides of the inside quadrilateral -- each parallel to one of the diagonals of ABCD -- are also perpendicular to each other. So, in fact, EFGH is a rectangle (a special kind of parallelogram).

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Problem C7

Using similar reasoning from Problem C6, you can prove that the inscribed quadrilateral must always be a parallelogram. In the diagram below, construct the diagonal BD. Using this diagonal as the base of two triangles (BDC and BDA), we have two triangles with midlines: FG is the midline of triangle BDC, and EH is the midline of triangle BDA. Each midline is parallel to the base, so FG and EH must be parallel to one another. By a similar argument (using diagonal AC), we can say that EF and HG must each be parallel to AC, and therefore parallel to one another. Then, definition 1 in Fact 3 tells us that EFGH must be a parallelogram, no matter how the original quadrilateral is constructed.

Note that for some concave quadrilaterals we can get the sides of the interior quadrilateral to coincide, so it's not really a parallelogram. Except in those rare cases, it will be a parallelogram.

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