Math 367-500
Solutions to HW#4
Fall 2011
1. Theorem 5 (section 4.5). Let H be a half plane with edge L, and let E be a plane that contains L but
not H. Then all points of H are on the same side of E.
Proof. Pick points P and Q in H. Suppose, P and Q lie on opposite sides of the plane E. Then, by
SS-1, P Q intersects the line L at the point R such that P − R − Q, which is impossible, because P and
Q lie on the same side of the line E. 2. Theorem 6 (section 4.5). The interior of the dihedral angle is always a convex set.
Proof. Let E1 and E2 be the planes that contain half planes H1 and H2 , respectively. Then the interior
of the dihedral angle D = H1 ∪ H2 ∪ L is the intersection of two sets:
(1) the side of E1 that contains H2
(2) the side of E2 that contains H1
By SS-1, sets in (1) and (2) are convex. By Theorem 8 (section 4.1), the interior of the dihedral angle is
convex. 3. Theorem 6 (section 4.6) If two intersecting lines form one right angle, then they form four right angles.
←→
←→
Proof. Let lines AB and CD intersect at the point P , and let ∠AP C is right, that is m∠AP C = 90.
Angles ∠AP C and ∠BP D form a vertical pair, therefore, by Vertical Angles Theorem, ∠AP C ∼
= ∠BP D,
therefore, m∠BP D = 90. Angles ∠AP C and ∠BP C form a linear pair, therefore, m∠AP C + m∠BP C =
180, Then, m∠BP C = 180 − m∠AP C = 90. Angles ∠BP C and ∠AP D form a vertical pair, therefore,
by Vertical Angles Theorem, ∠BP C ∼
= ∠AP D, therefore, m∠AP D = 90. 4. Write down all the congruences between an equilateral triangle and itself.
∼ BC, AB ∼
Solution. Let △ABC be an equilateral triangle. Then AB =
= AC, and BC ∼
= AC. In any
congruence between △ABC and itself,
AB ←→ AB, AB ←→ AC, and AB ←→ BC,
BC ←→ BC, BC ←→ AB, and BC ←→ AC,
AC ←→ AC, AC ←→ AB, and AC ←→ BC.
It follows that there are six congruences: △ABC ∼
= △ABC, △ABC ∼
= △ACB, △ABC ∼
= △BAC,
∼ △BCA, △ABC ∼
△ABC =
= △CAB, and △ABC ∼
= △CBA. 5. Given a triangle △ABC which is isosceles but not equilateral. That is, AB = AC, but AB 6= BC. How
many congruences are there, between △ABC and itself?
∼ AC but BC is not congruent to AB. In any congruence between △ABC
Solution. Suppose that AB =
and itself, the side AB will correspond then either to itself or to AC, the side AC will correspond either
to itself or to AB. The side BC can correspond only to itself. It follows that there are two congruences:
△ABC ∼
= △ABC and △ABC ∼
= △ACB. 1