This page has answers and/or notes about each question. The page does not have figures (which you are encourage to draw). Of course when answering such questions on a test, drawing and labeling a picture (not necessarily a careful construction) is almost always a good idea.
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Instructions: Do all problems. You will need straightedge and compass.
Prove that the perpendicular bisectors of the sides of a triangle ABC are concurrent.
Note: You can assume the basic axioms and theorems (except for theorems that are restatements of this theorem, e.g., the existence of a certain circle).
There is a proof in BG, Section 7.1. Also, this is the topic of Chapter 3 of GTC, especially Exp 3.3, Inv 1.
The proof uses the Perpendicular Bisector Locus Theorem (both directions), so it makes the proof easier to write if you state this theorem at the outset.
PBLT: For any two points A and B, the set of points equidistant from A and B is the perpendicular bisector of AB.
This theorem was Assignment 1. Now for the anwer to the problem.
Theorem: The perpendicular bisectors of a triangle ABC are concurrent.
Proof: Let P be the intersection of m, the perpendicular bisector of AB, and n, the perpendicular bisector of BC.
Then AP = BP since P is on m and BP = CP since P is on n (both by Perpendicular Bisector Locus Theorem).
Thus AP = BP = CP. But AP = CP implies that P is on the perpendicular bisector of AC, by PBLT.
This shows that all 3 perpendicular bisectors pass through P, so they are concurrent at P.
QED.
Let ABC be a triangle with right angle at C; point D is the foot of the altitude through C. If the lengths of the sides of ABC are a = |BC|, b = |CA|, c = |AB|, find the lengths of AD and CD in terms of a, b, and/or c. Show your reasoning.
Note: You can use any theorems.
Triangle ABC is similar to triangle ACD by AAS (shared angle A and right angles).
Therefore, AD/AC = AC/AB, or AD/b=b/c. So AD = b^2/c/
Also, CD/BC = AC/AB, or CD/a=b/c. So CD = ab/c.
Alternate method: Since ACD is similar to ABC and AC/AB = b/c, then the
ratio of similitude (or the scaling ratio) from ACD to ABC is b/c. Then the
triangle ABC with sides a, b, c is scaled to a triangle with these sides multiplied
by b/c, namely ab/c, b^2/c, b. Of course this is easier to follow with a sketch
or drawing.
Prove that a quadrilateral EFGH is a parallelogram if and only if its opposite angles are congruent (i.e., angle E = angle G and angle F = angle H).
Note: You can use any theorems except theorems about parallelograms.
The proof will use two theorems;
(1) If parallel lines are cut by a transversal, then the interior angles on the same side of the transversal are supplementary if and only if the two lines are parallel.
(2) The sum of the angles of a quadrilater = 360 degrees. (This can be quoted or proved by cutting the quadrilateral into 2 triangles.)
Proof;
Given EFGH with angle E = angle G and angle F = angle H.
Then angle E + angle G + angle F + angle H = 360 degrees, by (2).
Substitute to get 2(angle E + angle F)= 360 degrees, or angle E + angle F = 180 degrees.
Thus angle E and angle F are supplementary. These angles are interior angles formed by cutting lines EH and FG by the transversal EF. So by (1), the lines EH and FG are parallel.
Since angle F = angle H, then also angles E and H are supplementary, and the same argument shows that EF and HG are parallel.
So EFGH has opposite sides parallel. It is therefore a parallelogram by definition.
QED.
Construction: Given the triangle ABC in the figure, construct a circle tangent to all three sides (i.e., inscribed in the triangle).
The construction of the circle has two key steps. The center I of the circle is constructed as the intersection of any two of the bisectors of angles A, B, and C.
Then (don't skip this step!). The radius is constructed by constructing a perpendicular to one of the sides through I. If the foot of this perpendicular is F, then IF is a radius of the circle.
See GTC, Exp 5.3, Construction - Intersection of Angle Bisectors and the Incircle. Also BG, 7.3.
Construction: Given the segment DE in the figure, construct a point F so that DF/DE = 5/7.
See BG, page 54, figure 4.7. The only difference is that you construct a segment made of 7 congruent sub-segments instead of 5.
