Math 367-500
Solutions to HW#3
Fall 2011
1. Given a triangle, and a line lying in the same plane. If the line intersects the interior of the triangle, then
it intersects at least one of the sides of the triangle.
Proof. Let us prove contra positive which is equivalent. Given a triangle, and a line lying in the same
plane. If the line does not intersect any of the sides of the triangle, then it does not intersect the interior
of the triangle.
If L does not intersect sides of the triangle △ABC, then its vertices A, B, and C lie on the same side of
L, which we denote by H. By SS-1, H is convex, therefore AB, BC, and AC lie in H. Pick an arbitrary
point D in the interior of △ABC. By Theorem 9 (section 4.2), D lies in the interior of ∠BAC. Then,
−−→
by the Crossbar Theorem, the ray AD intersects BC at a point P such that B − P − C. P lies on BC,
therefore P is in H. Since H is convex, then AP lies in H, therefore D ∈ H. The interior of the triangle
lies in H, so L does not intersect it. 2. Show that if the diagonals of a quadrilateral intersect each other, then the quadrilateral is convex.
Proof. Let us prove contra positive which is equivalent. If a quadrilateral is not convex, then its diagonals
do not intersect.
Assume that ABCD with diagonals AC and BD is not convex. Then one of its sides does not lie if the
half plane determined by the opposite side. Up to a relabeling, we can assume that side AD does not lie
←→
on the half plane determined by BC. That is, points A and D lie in opposite sides of the line BC. Then,
by Theorem 1 (section 4.3), AC does not intersect BD. 3. Show that if each vertex of a quadrilateral lies in the interior of the opposite angle, then the quadrilateral
is convex.
Proof. A lies in the interior of ∠BCD, then, by the definition of the interior of an angle,
←→
(i.) A and B lie on the same side of DC
←→
(ii.) A and D lie on the same side of BC
(i.) = condition (1) from the definition of a convex quadrilateral
(ii.) = condition (1) from the definition of a convex quadrilateral
B lies in the interior of ∠ADC, then
←→
(iii.) A and B lie on the same side of DC
←→
(iv.) B and C lie on the same side of AD
(iii.) = condition (1) from the definition of a convex quadrilateral
(iv.) = condition (2) from the definition of a convex quadrilateral
C lies in the interior of ∠DAB, then
←→
(v.) C and B lie on the same side of AD
←→
(vi.) C and D lie on the same side of AB
(v.) = condition (2) from the definition of a convex quadrilateral
(vi.) = condition (3) from the definition of a convex quadrilateral
D lies in the interior of ∠ABC, then
←→
(vii.) D and A lie on the same side of BC
←→
(viii.) C and D lie on the same side of AB
(vii.) = condition (2) from the definition of a convex quadrilateral
(viii.) = condition (3) from the definition of a convex quadrilateral
Since ABCD satisfy conditions (1)-(4) from the definition of a convex quadrilateral, then ABCD is
convex. 1
4. Show that for any quadrilateral (convex or not), the lines containing the diagonals always intersect.
Proof. If ABCD is convex, then, by Theorem 1, its diagonals always intersect.
If ABCD is not convex, then by Problem 2, its diagonals AC and BD do not intersect, therefore points
←→
A and C lie on the same side of BD.
←→
←→
C and B lie on the same side of AD. and C and D lie on the same side of AB, therefore, C lies in the
−→
←→
interior of ∠BAD. By the Crossbar Theorem, AC intersects BD at a points Q such that B − Q − D.
←→
←→
Thus, AC intersects BD. 2