section12

ANGLES AND PARALLEL LINES, TRIANGLE'S MIDSEGMENTS.

In order to investigate the further properties of quadrilaterals first we have to discuss some results involving angles and parallel lines.

Theorem 12.1 Consider an angle and parallel lines that intersect with the angle. If the parallel lines divide one side of the angle into equal parts then they divide the other side of the angle into equal parts as well.

Proof Let's consider an angle P1BQ1 and three parallel lines m1, m2, and m3 that intersect with P1BQ1.

Suppose that P1P2 = P2P3 then we need to show that Q1Q2 = Q2Q3. Let EF be a line that passes through Q2 and is parallel to P1B. Then we have two parallelograms: P3FQ2P2 and P2Q2EP1. From the theorem 11.1 it follows that P2P3 = FQ2 and P1P2 = Q2E. But P1P2 = P2P3 by the assumption of our theorem, therefore

Triangles Q3FQ2 and Q1Q2E are equal by ASA: angles Q3Q2F and Q1Q2E are equal as vertical angles, angles Q3FQ2 and Q1EQ2 are equal as alternate interior angles for parallel lines m3 and m1 and their transversal EF, FQ2 = EQ2 from above. Therefore Q1Q2 = Q2Q3.

The case when we have more than three parallel lines is proved in the same way by considering each consecutive triple of parallel lines.

Corollary Suppose that a and b are two arbitrary lines and m1, m2, m3 etc. are parallel lines that intersect with a and b. If m1, m2, m3 etc. divide one of the lines a and b into equal parts then they divide the second line into equal parts as well.

With such a useful result as the theorem 12.1 we are going to return for a moment to our discussion of triangles and add some new facts.
Consider an arbitrary triangle ABC and let P, Q and R be the middle points of sides AB, BC and AC respectively. Then segments PQ, QR and PR are called the triangle's midsegments.

Theorem 12.2 Consider an arbitrary triangle ABC and any of its midsegments that connects two sides of ABC. Then the midsegment is parallel to the third side and the length of the midsegment is equal to the half of the length of the third side.

Proof Let's prove the theorem for the midsegment PQ in the picture below. We need to show that PQ is parallel to AC and that the length of PQ is equal to the half of the length of AC.

Let m be the line parallel to AC and passing through Q. It intersects with AB at some point S. BQ = QC from the assumption of the theorem, m and AC are parallel lines, therefore from the theorem 12.1 BS = SA. But we have already assumed that BP = PA, so points S and P coincide and PQ is parallel to AC.

Let QR be one more midsegment of the triangle ABC. By the same reasoning as above QR is parallel to AB. Then PQRA is a parallelogram with PQ = AR.
We also know that AR = RC (since we assumed that R is the middle point of AC). Therefore

EXERCISE SET #12

Problem 1. The perimeter of a triangle is the sum of the lengths of all triangle's sides.

Suppose that ABC is a triangle with AB = 8, BC = 10 and AC = 12 inches. If Q is the middle point of AB, P is the middle point of BC and R is the middle point of AC, find the perimeter of the triangle PQR .

Problem 2. Suppose that ABC is an isosceles triangle with the perimeter equal to 16 inches. If a midsegment of ABC is equal to 3 inches, find the lengths of the triangle's sides. Hint: consider all possible cases.