Unit 3: Parallel Lines, Proportional Segments and Angle Bisectors

• Theorem: a straight line drawn parallel to one side of a triangle, cutting the other two sides, divides them in the same ratio: if DE ∥ BC with D on AB and E on AC, then AD/DB = AE/EC.
• Equivalent forms that follow from it: AD/AB = AE/AC = DE/BC — which is the same as the similarity △ADE ~ △ABC.
• Use the first form when the question involves the two parts of a side, and the second when it involves a whole side or the parallel segment itself.

• Converse: if a point D on AB and a point E on AC divide those sides proportionally (AD/DB = AE/EC), then DE ∥ BC.
• Method: compute the two ratios separately, simplify, and compare. Equality proves parallelism; inequality refutes it.
• Always check that D and E lie on the sides (or their extensions consistently) before applying.

• Putting D and E as the midpoints of AB and AC gives AD/DB = 1 = AE/EC, so by the converse DE ∥ BC.
• Moreover DE/BC = AD/AB = 1/2, so: the segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it.
• The three midpoints of a triangle divide it into four congruent triangles, each similar to the original with scale factor 1/2.
• Corollary: the line through the midpoint of one side of a triangle, drawn parallel to a second side, bisects the third side.

• A set of parallel lines cuts equal intercepts on any transversal, therefore it cuts equal intercepts on every other transversal.
• General form: if three parallel lines cut two transversals at A, B, C and at D, E, F respectively, then AB/BC = DE/EF, and likewise AB/DF = BC/EF … etc.
• Construction aid: to divide a segment into n equal parts, step off n equal units on an auxiliary ray and draw the parallel through the last point.
• Equally spaced parallel lines preserve ratios of intercepted lengths, which is how a scaled ruler or a grid is divided proportionally.

• In trapezoid ABCD with AB ∥ CD, a line XY parallel to the two bases with X on AD and Y on BC gives AX/XD = BY/YC.
• The segment joining the midpoints of the two legs (the median) is parallel to the bases and equals half their sum: median = (AB + CD)/2.
• The portion of such a parallel between the legs, when X divides AD in the ratio m : n, has length (m·CD + n·AB)/(m + n).
• Lines through the intersection point of the diagonals give proportional segments on each diagonal, used to prove that the diagonal intersection divides them in the ratio of the bases.

• Theorem: if AD bisects angle BAC of triangle ABC and D lies on BC, then BD/DC = AB/AC — the bisector divides the opposite side in the ratio of the two sides forming the angle.
• Mnemonic: the ratio in which the bisector cuts the opposite side equals the ratio of the adjacent sides.
• Converse: if D on BC satisfies BD/DC = AB/AC, then AD bisects angle BAC.
• Rearrangements used in exams: BD = BC·AB/(AB + AC), DC = BC·AC/(AB + AC), and BD/BC = AB/(AB + AC).

• The bisector of the exterior angle at A meets the extension of BC at a point E such that BE/EC = AB/AC, with E outside the segment BC, on the side of the shorter adjacent side.
• If AB = AC the external bisector is parallel to BC, so it never meets the line BC and there is no external division point.
• Converse: verify that E lies on the extension of BC (outside the segment) and that BE/EC = AB/AC; then AE is the external bisector.
• The internal and external bisectors at the same vertex are perpendicular to each other, so D and E subtend a right angle at A; B, D, C, E form a harmonic range with AB/AC.

• Internal: AD² = AB × AC − BD × DC.
• External: AE² = BE × EC − AB × AC (E on the extension of BC).

Check on a 3-4-5 triangle

• Take AB = 3, AC = 4, BC = 5. The internal bisector gives BD = 15/7 and DC = 20/7, so AD² = 12 − 300/49 = 288/49 → AD = 12√2/7.
• The external bisector gives BE = 15, EC = 20, so AE² = 300 − 12 = 288 → AE = 12√2 — note AE > AD, as it must be.

• Triangles ABD and ACD share the same altitude from A, so area(△ABD)/area(△ACD) = BD/DC; using the bisector theorem this ratio is also AB/AC.
• Combining with similarity: a ratio of lengths gives a ratio of areas only when the triangles are similar or share an altitude — state which one you are using.
• Perimeter problems: if a bisector or a parallel divides a side, first find the two parts with the ratio, then substitute into the length formulas.
• Coordinate setups: assign the base on the x-axis and use the section formula with the ratio from the bisector theorem to locate D.