Chapter 8.2 — Exercise 8.1 — BPT Problems

Proving a Triangle Isosceles From a Ratio

In △PQR, ST is drawn so that PS/SQ = PT/TR, and it's also given that ∠PST = ∠PRQ. The goal: prove △PQR is isosceles.

The ratio condition PS/SQ = PT/TR is exactly the converse of BPT's trigger — it forces ST ∥ QR immediately. Once ST ∥ QR is established, PQ becomes a transversal cutting two parallel lines, so ∠PST = ∠PQR (corresponding angles). Combined with the given ∠PST = ∠PRQ, that gives ∠PQR = ∠PRQ — and a triangle with two equal angles always has the two sides opposite those angles equal too, so PQ = PR. △PQR is isosceles. Every step here is either "apply the converse of BPT" or "use a parallel-line angle fact" — nothing else.

Two Proofs, One Shared Shape

Two problems in this exercise share an identical structure: a single point sits at the apex of two triangles that share one side, a parallel line is drawn in each triangle separately, and BPT applied to both triangles combines into one final ratio.

Both proofs follow the identical two-step shape: apply BPT once in each of the two triangles that share a vertex, and since both results equal the same middle ratio (AL/AC in the first row, BD/DA in the second), they must equal each other. Neither proof needs anything beyond BPT itself, applied twice instead of once. Spotting the shared middle ratio before writing anything down is really the whole skill here — once it's clear that both applications of BPT are secretly computing the same quantity from two different triangles, the final equality is just a matter of substitution.

The Midpoint Theorem, Both Directions

Two more problems form a genuine pair — one is the direct statement, the other is its converse, and each one is really BPT (or its converse) applied to the single special case where a ratio equals exactly 1.

• A line through one midpoint, parallel to a second side, bisects the third: let D be the midpoint of AB, so AD/DB=1. Drawing DE ∥ BC through D, BPT gives AD/DB = AE/EC — and since AD/DB=1, that forces AE=EC, meaning E is also a midpoint. DE bisects AC.
• The segment joining two midpoints is parallel to the third side (the converse direction): let D and E be the midpoints of AB and AC. Then AD/DB=1 and AE/EC=1 automatically, so AD/DB=AE/EC — and by the converse of BPT, that ratio equality alone is enough to conclude DE ∥ BC.

Both proofs are worth noticing as mirror images of each other: the first assumes a parallel line and proves a ratio of 1; the second assumes a ratio of 1 and proves a parallel line. Swapping which fact is "given" and which is "to prove" is the only thing separating them.

Chaining Through a Common Vertex, Twice Over

Two further problems chain BPT through two triangles and then invoke the converse to finish — a three-step proof instead of the two-step pattern used above.

• Rays from O through P, Q, R, with points A, B, C on them: given AB ∥ PQ and AC ∥ PR, show BC ∥ QR. In △OPQ, AB∥PQ gives OA/AP = OB/BQ. In △OPR, AC∥PR gives OA/AP = OC/CR. Both equal the same ratio OA/AP, so OB/BQ = OC/CR — and by the converse of BPT, BC ∥ QR.
• A parallel setup inside two triangles sharing a vertex and a side: given DE ∥ OQ and DF ∥ OR, show EF ∥ QR. In △POQ, DE∥OQ gives PE/EQ = PD/DO. In △POR, DF∥OR gives PD/DO = PF/FR. Both equal PD/DO, so PE/EQ = PF/FR — and again by the converse of BPT, EF ∥ QR.

Structurally these two proofs are identical to each other, just with different vertex labels — BPT applied once in each of two triangles sharing a common vertex and a common ray, followed by one application of the converse to close the proof. Recognising this shared shape is more useful than memorising either proof individually, since a new problem built the same way is solved the same way regardless of what its points happen to be called.

A Trapezium's Diagonals

ABCD is a trapezium with AB ∥ DC, and its diagonals meet at O. The goal: show AO/BO = CO/DO. Since a trapezium's diagonals don't sit inside a single triangle on their own, the proof needs one auxiliary line: draw EO through O, parallel to both AB and DC, meeting AD at E.

Both results share the same middle ratio AE/ED, so AO/OC = BO/OD — which rearranges directly into AO/BO = CO/DO. The auxiliary parallel line is the one genuinely new idea in this proof compared to every problem before it: instead of applying BPT inside triangles that already exist in the figure, this proof first builds a triangle-friendly line to apply BPT to at all. Recognising when a figure needs this kind of construction — a quadrilateral instead of a triangle, a diagonal instead of a side — is a skill on its own, separate from applying BPT correctly once the right triangle actually exists to apply it to.

A Construction, Not a Proof

The exercise's final problem breaks from the proof pattern entirely: draw a 7.2 cm line segment and divide it in the ratio 5:3 using compass and straightedge, then measure the two resulting parts. Dividing 7.2 cm in ratio 5:3 means splitting it into 8 equal shares of 0.9 cm each — 5 shares make one part, 3 shares make the other, giving parts of 4.5 cm and 2.7 cm. The actual construction itself uses a completely different toolkit (drawing a ray at an angle, marking off 8 equal arcs with a compass, and joining back to the original segment) than anything else in this exercise — proof and construction are related but genuinely separate skills.

From Ratios to Criteria

Every proof in this exercise treated "similar enough to share a ratio" as something already given by a parallel line. Exercise 8.2 asks the question this exercise never had to: what exactly makes two entire triangles similar to begin with, and how many angles or sides actually need to be checked before that conclusion is safe to draw. For the theorem and converse every proof above depends on, revisit the chapter introduction.