Geometry: Similarity, Pythagoras, Circle and Coordinates
Geometry (Mathematics Part II) tests deductive reasoning rather than computation. Marks are given for naming the theorem or test used and for writing correspondence in the correct order; an answer with the right number but wrong correspondence still loses marks.
Similarity
Triangles are similar by AA (two pairs of equal angles), SAS (two sides in proportion with the included angle equal) and SSS (all three sides in proportion). If ΔABC ~ ΔPQR then AB/PQ = BC/QR = CA/RP and ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R. Areas follow A(ΔABC)/A(ΔPQR) = (AB/PQ)². The Basic Proportionality Theorem states that in ΔABC a line parallel to BC meeting AB at D and AC at E gives AD/DB = AE/EC, and its converse proves the line parallel.
Pythagoras and Apollonius
In a right triangle, hypotenuse² = sum of the squares of the other two sides; the converse identifies a right angle from the three side lengths. Apollonius: if M is the midpoint of BC in ΔABC, then AB² + AC² = 2AM² + 2BM².
Circle
A tangent is perpendicular to the radius at the point of contact; tangent segments drawn from one external point are equal; the angle between a tangent and a chord equals the inscribed angle in the alternate segment; opposite angles of a cyclic quadrilateral are supplementary.
Coordinate geometry
Distance = √[(x₂ − x₁)² + (y₂ − y₁)²]; midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2); section formula for ratio m : n gives ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)).
Worked example
In ΔABC, DE ∥ BC with AD = 4, DB = 6, AE = 5. By BPT, AD/DB = AE/EC, so 4/6 = 5/EC and EC = 30/4 = 7.5. Then A(ΔADE)/A(ΔABC) = (AD/AB)² = (4/10)² = 16/100 = 4/25.
Traps that cost marks
Mismatching vertices in the similarity statement, using the side ratio instead of its square for areas, substituting AB/BC into BPT, and assuming a triangle is right-angled without testing the converse are the usual losses.