Parallelograms And Trapezoids

page revision: 0, last edited: 15 Sep 2008 08:58

Parallelograms And Trapezoids

page revision: 0, last edited: 15 Sep 2008 08:58

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MN=NQ should be MP=NQ

ReplyOptionsPage 71, line 4 after the pictures: "(since AO=OD), and so are B and C" should be "(since AO=OC), and so are B and D".

In the next sentence, "extension of line" should better be "extension of this line".

On page 74, line 1: "DL" should be "EL"

ReplyOptionsExercise 174:

Is a parallelogram considered a trapezoid.No, $\S 84$) a parallelogram is a quadrilateral with pairwise parallel sides. $\S 96$) a trapezoid has two opposite sides that are parallel and the other two sides are not parallel.

ReplyOptionsExercise 175:

How many centers of symmetry can a polygon have?One. Any two points of the polygon can only be along one line. $\S 88$) the midpoint of this line is the point of symmetry.

ReplyOptionsYour argument is incorrect. Counter-example: any centrally symmetric polygon, e.g. a square ABCD. The line connecting vertices A and B is not bisected by the center of symmetry.

ReplyOptionsExercise 176:

Can a polygon have two parallel axes of symmetry?No. For any two symmetric points, they can both only exist along one straight line, $\S 4$.

Axial symmetry, $\S 37$, requires the symmetric points to be along the same perpendicular to the axis of symmetry. Thus this line through the two symmetric points must be perpendicular to the axis of symmetry.

The foot of this perpendicular along the axis of symmetry is unique, $\S 23$, so there is only one such perpendicular.

If there were a second axis of symmetry it must be the same distance away from the points and perpendicular to the line between the symmetric points. It must then be parallel to the first axis of symmetry and through the same points, but $\S 75$ parallel postulate states there is only one such line.

ReplyOptionsYour argument seems incorrect again. Since you never use the fact that the figure is a polygon, the reasoning should be applicable to any figure. However there exist figures (for example, the whole plane) with two (or in fact any number) of parallel axes of symmetry.

ReplyOptionsExercise 177:

How many axes of symmetry can a quadrilateral have?$\S 92$) the square has four axes of symmetry.

ReplyOptionsExercise 178:

Midpoints of the sides of a quadrilateral are the vertices of a parallelogram. Determine under what conditions this parallelogram will be (a) a rectangle, (b) a rhombus, (c) a square.$\S 92$) square is a rhombus with all angles right. $\S 91(1)$ Rhombus diagonals are perpendicular and bisect the angles at the vertices.

Connect the midpoints of the given quadrilateral by lines. Connect the vertices of the vertices of the parallelogram just constructed via lines.

If these diagonals are perpendicular and half of the first diagonal is congruent to half of the second diagonal, the four triangles these diagonals divide the parallelogram into are congruent, $\S 40(1) SAS$. $\S 81(4)$ each acute angle in an isoceles right triangle is $\frac{1}{2}d$. Thus, the angles in the parallelogram are right, and it is a square.

If the diagonals are only perpendicular and not congruent, it is a rhombus.

If the diagonals are congruent and not perpendicular, it is a rectange, $\S 90(1)$.

ReplyOptionsExercise 179:

In a right triangle, the median to the hypotenuse is congruent to half of it.Let the right triangle be $\triangle ABC$. Construct the midline $EF$ parallel to base $BC$, $\S 95$.

$AE = BE$, $\angle AEF = \angle ABC = d$, $\angle AEF = \angle BEF$, $\S 22$) supplementary angles.

$\triangle AEF = \triangle BEF$, $\S 40(1) SAS$. $AF = BF$.

ReplyOptionsAlternatively, we could have drawn a median from E to a point K along AF, and also a median from E to a point K' along BF. By the SAS-test, triangle AEF is congruent to triangle EBF, so the medians (EK and EK') are also congruent by the ASA-test (since congruence of AEF and EBF implies congruence of angles EFA and EFB, EF is common, and the other sides enclosing those angles (FK and EK') are congruent as halves of respectively congruent sides). Therefore, since E is the midpoint of AB by construction, and K' is also a midpoint of FB, EK = EK' = half of AF by the midline theorem.

ReplyOptionsExercise 180:

Conversely, if a median is congruent to a half of the side it bisects, then the triangle is right.Let the triangle be $\triangle ABC$, median to $AC$ is $BD$. Given $BD = AD = CD$.

Connect the midpoint of $AB$, $E$, to the median point $D$. $ED$ is the midline and parallel to the base $BC$, $\S 95$.

$\triangle ADB$ is isoceles, $\S 33$, and the base angles, $\angle DAE$ and $\angle DBE$, are congruent, $\S 35(2)$.

$\triangle EAD = \triangle EBD$ by $\S 40(1) SAS$. $\angle AED = \angle BED$, but these are supplementary angles ($\S 22$) and must be each congruent to $d$.

$ED$ is perpendicular to $AB$ and also parallel to $BC$. $\S 73(3)$ - $\angle DEB$ and $\angle CBE$ are same side interior angles whose sum is $2d$.

$\angle CBE = d$, $\triangle ABC$ is right, $\S 33$.

ReplyOptionsExercise 181:

In a right triangle, the median and the altitude drawn to the hypotenuse make an angle congruent to the difference of the acute angles of the triangle.In the isoceles right triangle, the median and altitude are one in the same, so this angle would be 0. In this case the acute angle different would be: $\frac{1}{2}d - \frac{1}{2}d = 0$.

Let the right triangle be $\triangle ABC$, median $BD$, and altitude $BE$. Angle made by median and altitude is $\angle DBE$.

$\S 81$) Angle sum of a triangle is $2d$. Acute angles $\angle BAC = \angle BCA = d$ because the third angle is right.

#179) $BD = AD = CD$. $\triangle ADB is isoceles, [[$ \S 33$, and its base angles are congruent, $\angle DAB = \angle DBA$, $\S 35(2)$.

$\triangle BDC$ is isoceles, $\angle DBC = \angle DCB$.

$\triangle BEC$ is right as given. $\angle EBC = d - \angle ECB$. $\angle BCA = \angle ECB = \angle DCB$. $\angle BAC = d - \angle BCA$.

Thus, $\angle EBC = \angle BAC$. $\angle DBC = \angle DBE (the angle between the median and altitude) + \angle EBC$.

$\angle DBC - \angle EBC = \angle DBE$, which is the same as $\angle DCB - \angle BAC = \angle DBE$ and what was to be proved.

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