IGCSE, Edexcel, IB and AP Science Courses | Dr Dabour Hub Academy

Overview

State and prove the theorem of the line parallel to one side of a triangle, and use its converse to establish parallelism.
Use Thales’ (intercept) theorem and its consequences in the triangle and the trapezoid.
Apply the internal and external bisector theorems, their converses, and the formulas for the lengths of the two bisectors.

Parallel Lines and Proportional Segments
├── 3-1 The line parallel to one side of a triangle
│   ├── the theorem and other ratios
│   ├── the converse
│   ├── the midline (joining two midpoints)
│   ├── Thales' general theorem (equal intercepts)
│   └── applications in the trapezoid
└── 3-2 The angle bisectors
    ├── internal bisector theorem and converse
    ├── external bisector theorem and converse
    ├── special case AB = AC
    ├── length of the internal and external bisectors
    └── mixed applications with areas

Unit 3: Parallel Lines, Proportional Segments and Angle Bisectors · 1 / 11

3-1 The line parallel to one side of a triangle

Comparison aspect	Main idea	Keywords
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	straightlinedrawnparallel
Equivalent forms that follow from it	AD/AB = AE/AC = DE/BC — is the same as the similarity △ADE ~ △ABC	samesimilarity
•	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	firstformquestioninvolves

Unit 3: Parallel Lines, Proportional Segments and Angle Bisectors · 2 / 11

3-2 The converse

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

Unit 3: Parallel Lines, Proportional Segments and Angle Bisectors · 3 / 11

3-3 Consequences: the midline

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.

Unit 3: Parallel Lines, Proportional Segments and Angle Bisectors · 4 / 11

3-4 Thales’ general theorem (the intercept theorem)

Parallel Lines, Proportional Segments: three parallel shelves cutting two leaning rods show parallel lines and proportional segments: 3/2 = 6/4.

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.

Unit 3: Parallel Lines, Proportional Segments and Angle Bisectors · 5 / 11

3-5 Applications in the trapezoid

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.

Unit 3: Parallel Lines, Proportional Segments and Angle Bisectors · 6 / 11

3-6 The internal bisector theorem and its converse

Comparison aspect	Main idea	Keywords
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	bisectsangletrianglelies
Mnemonic	the ratio in the bisector cuts the opposite side equals the ratio of the adjacent sides	ratiobisectorcutsopposite
Converse	if D on BC satisfies BD/DC = AB/AC then AD bisects angle BAC	satisfiesthenbisectsangle
Rearrangements used in exams	BD = BC·AB/(AB + AC) DC = BC·AC/(AB + AC) and BD/BC = AB/(AB + AC)

Unit 3: Parallel Lines, Proportional Segments and Angle Bisectors · 7 / 11

3-7 The external bisector theorem

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.

Unit 3: Parallel Lines, Proportional Segments and Angle Bisectors · 8 / 11

3-8 Lengths of the internal and external bisectors

InternalAD² = AB × AC − BD × DC.

ExternalAE² = BE × EC − AB × AC (E on the extension of BC).

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.

Unit 3: Parallel Lines, Proportional Segments and Angle Bisectors · 9 / 11

3-9 Mixed applications with areas

Comparison aspect	Main idea	Keywords
•	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	Trianglessharesamealtitude
Combining with similarity	a ratio of lengths gives a ratio of areas only when the triangles are similar or share an altitude — state one you are using	ratiolengthsgivesareas
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	bisectorparalleldividesside
Coordinate setups	assign the base on the x-axis and use the section formula with the ratio from the bisector theorem to locate D	assignbaseaxissection

Unit 3: Parallel Lines, Proportional Segments and Angle Bisectors · 10 / 11

Exam traps and key takeaways

💡 High-yield traps

AD/DB = AE/EC uses the parts; AD/AB = AE/AC uses the whole sides. Mixing them is the standard lost mark.
The converse needs the equality of the two ratios before you may write “therefore parallel”.
The midline is half the third side — do not double it by accident when solving for the base.
The internal bisector ratio is AB/AC in the order of the sides: BD (adjacent to AB) corresponds to AB.
The external bisector divides the base externally; if you get a point inside the segment you have used the internal case.
When AB = AC there is no external division point because the external bisector is parallel to BC.
Subtract in the internal formula, subtract the other way in the external formula: AD² = AB·AC − BD·DC; AE² = BE·EC − AB·AC.
Area ratio = base ratio only when the two triangles share the same altitude.

Unit 3: Parallel Lines, Proportional Segments and Angle Bisectors · 11 / 11

Curriculum

Mathematics — First Secondary

Unit 1

Unit 2

Unit 3

Unit 4

Unit 3: Parallel Lines, Proportional Segments and Angle Bisectors