MATH 241 Winter 2013
Assignment 5
Due Friday February 15 by 12:00 noon
in the assignment box on the 3rd floor of CAB
(1) (6 points) Question 2.7.1 from the text: Let BE and CF be altitudes
of 4ABC, and M the midpoint of BC. Show that M E = M F .
We recognize that ∠BF C and ∠BF C are 90◦ and thus by Thales’
Theorem both E and F are points on a semicircle with base BC as
in the diagram below.
A
F
B
E
M
C
As M is the midpoint of BC, M is the center of this circle, and hence
M E = M F as they are both radii of the circle.
(2) (7 points) Question 2.7.2 from the text: BE CF are altitudes of
4ABC, and EF is parallel to BC. Prove that 4ABC is isosceles.
As ∠BF C = ∠BF C = 90◦ , by Thales’ Theorem E and F are
points on a semicircle with base BC as in the diagram.
A
F
E
B
C
Now we have
∠F BE = ∠F CE Thales’ Theorem
∠EBC = ∠EF C Thales’ Theorem
= ∠F CB as F E k BC
1
2
So
∠ABC = ∠F BE + EBC
= ∠F CE + ∠F CB
= ∠ACB
By the converse of the isosceles triangle theorem we have that AB =
AC and that 4ABC is isosceles.
(3) (7 points) Question 2.7.16 from the text: ABCD is a nonsimple
quadrilateral. P, Q, R and S are the midpoints of AB, BC, CD, and
DA respectively. Show that P QRS is a parallelogram.
Consider the diagram below.
S
A
D
P
R
B
C
Q
By the Midline Theorem, segments P Q = SR = AC/2 and the three
segments are parallel. Similarly, QR = P S = BD/2 and these three
segments are parallel. So P QRS is a parallelogram.
3
(4) (10 points) Construct a triangle ABC with BC below, so that the
altitude from B has the length of the segment hb that is given, and
so that the median from A has the length of ma as given. Construct
this on this piece of paper! Do not adjust the scale or position of BC
on this page.
A
A
hb
B
M
C
A
A
Solution
• Construct the perpendicular bisector of BC to find M , the midpoint to BC. And use M to construct the circle C(M, ma ). The
point A must be on this circle.
• Construct C(B, hb ). The side CA must be tangent to this circle.
• Construct a line l that passes through C and is tangent to
C(B, hb ). (There are two such lines - l 1 and l 2 .)
• Let A be one of the points of intersection of C(M, ma ) with l
and connect A to B and C.
The 4ABC has median through A of length ma since A is on
C(M, ma ) and it’s altitude from B has length hb since CA is tangent
to C(B, hb ).
ma