3.1 Plane Geometry: Angles, Triangles, Polygons, and Circles

3.1 Plane Geometry: Angles, Triangles, Polygons, and Circles

Plane geometry on the AdUCET rewards recognition, not derivation. A typical item hands you a figure or a one-sentence description and expects you to name the single relationship that unlocks it, quickly and without a diagram of your own. Every problem in this section is designed to be finished by hand, so work in exact form: leave $\pi$ and surds such as $\sqrt{3}$ unresolved unless the answer choices are decimals.

1. Angles, linear pairs, and the transversal family

• Complementary angles sum to $90^\circ$; supplementary angles sum to $180^\circ$. An angle of $38^\circ$ has complement $52^\circ$ and supplement $142^\circ$.
• A linear pair — two adjacent angles whose outer sides form a straight line — is always supplementary.
• Vertical angles, the opposite pair produced when two lines cross, are always congruent.
• Angle bisector: a ray that splits an angle into two congruent halves.

When a transversal (a line crossing two others) cuts a pair of parallel lines, eight angles appear but only two distinct measures exist: some $x$ and its supplement $180^\circ - x$.

                    \   transversal T
                1    \   2
L1 -------------------\-----------------
                 3     \  4
                        \
                         \
                    5     \  6
L2 ------------------------\------------
                      7     \  8
                             \

The converse is examinable too: if one pair of corresponding angles is congruent, the two lines must be parallel. That is how items ask you to justify parallelism rather than compute a number.

2. Triangles

• Angle Sum Theorem: the three interior angles of any triangle total $180^\circ$.
• Exterior Angle Theorem: an exterior angle equals the sum of the two remote (non-adjacent) interior angles, so it is always larger than either one.
• Triangle Inequality: the sum of any two sides must exceed the third. Equivalently, with sides $a$ and $b$ fixed, the third side $c$ satisfies $|a-b| < c < a+b$. Sides $7$ and $10$ admit any third side strictly between $3$ and $17$.
• Side–angle correspondence: the longest side lies opposite the largest angle.
• An isosceles triangle has congruent base angles, and the converse holds; an equilateral triangle is equiangular at $60^\circ$ each, with area $A = \frac{\sqrt{3}}{4}s^2$.

Congruence and similarity criteria

Congruent triangles are identical in size and shape. The valid shortcuts are SSS, SAS, ASA, AAS, plus HL (hypotenuse–leg) for right triangles only. SSA is ambiguous and AAA fixes only shape, so neither proves congruence.

Similar triangles have equal angles and proportional sides. Prove similarity by AA (two pairs of equal angles), SAS similarity (one equal angle between proportional sides), or SSS similarity (all three ratios equal). If the linear scale factor is $k$, then corresponding perimeters are in ratio $k$ but corresponding areas are in ratio $k^2$.

The Basic Proportionality (Thales) Theorem is the workhorse: a line parallel to one side of a triangle cuts the other two sides proportionally.

Special right triangles and Pythagorean triples

$$a^2 + b^2 = c^2$$

Memorize the triples $3$-$4$-$5$, $5$-$12$-$13$, $8$-$15$-$17$, $7$-$24$-$25$, $9$-$40$-$41$ and their multiples ($6$-$8$-$10$, $9$-$12$-$15$, $10$-$24$-$26$, $15$-$20$-$25$, $21$-$28$-$35$). Spotting one saves you a square-root extraction.

3. Quadrilaterals and polygons

1. Trapezoid: exactly one pair of parallel sides; the median joining the leg midpoints equals $\frac{1}{2}(b_1+b_2)$. An isosceles trapezoid has congruent legs and congruent base angles.
2. Parallelogram: both pairs of opposite sides parallel and congruent, opposite angles congruent, consecutive angles supplementary, diagonals bisect each other.
3. Rectangle: a parallelogram with four right angles; its diagonals are congruent.
4. Rhombus: a parallelogram with four congruent sides; its diagonals are perpendicular and bisect the vertex angles.
5. Square: simultaneously a rectangle and a rhombus.

For a convex polygon with $n$ sides: interior angles total $(n-2)\times 180^\circ$; exterior angles always total $360^\circ$ no matter how many sides; a regular $n$-gon has each interior angle $\frac{(n-2)180^\circ}{n}$ and each exterior angle $\frac{360^\circ}{n}$; the number of diagonals is $\frac{n(n-3)}{2}$.

4. Circles

• Circumference $C = 2\pi r = \pi d$; area $A = \pi r^2$.
• Arc length for central angle $\theta$ in degrees: $L = \frac{\theta}{360^\circ}\cdot 2\pi r$.
• Sector area: $A_{\text{sec}} = \frac{\theta}{360^\circ}\cdot \pi r^2$. A segment is the sector minus the triangle formed by the two radii.
• Inscribed Angle Theorem: an inscribed angle is half the central angle subtending the same arc; consequently an angle inscribed in a semicircle is exactly $90^\circ$, inscribed angles on the same arc are congruent, and opposite angles of a cyclic quadrilateral are supplementary.
• Tangent–radius: a tangent is perpendicular to the radius at the point of tangency, and the two tangent segments drawn from one external point are congruent.
• A radius perpendicular to a chord bisects that chord.

5. Worked example 1 — parallel cut inside a triangle

Problem. In $\triangle ABC$, point $D$ lies on $\overline{AB}$ and $E$ on $\overline{AC}$ with $\overline{DE}\parallel\overline{BC}$. Given $AD = 6$ cm, $DB = 4$ cm, $AE = 9$ cm and $BC = 15$ cm, find (a) $EC$, (b) $DE$, (c) the ratio of the area of $\triangle ADE$ to the area of trapezoid $DBCE$.

1. Proportionality. $\dfrac{AD}{DB} = \dfrac{AE}{EC} \Rightarrow \dfrac{6}{4} = \dfrac{9}{EC} \Rightarrow EC = \dfrac{4\times 9}{6} = 6$ cm.
2. Similarity. $\angle A$ is shared and $\angle ADE = \angle ABC$ (corresponding angles), so $\triangle ADE \sim \triangle ABC$ by AA. The scale factor uses whole sides: $k = \dfrac{AD}{AB} = \dfrac{6}{6+4} = \dfrac{3}{5}$.
3. $DE = k\cdot BC = \dfrac{3}{5}(15) = 9$ cm.
4. Areas. $\dfrac{[ADE]}{[ABC]} = k^2 = \dfrac{9}{25}$, so the trapezoid takes the remaining $16$ parts and $[ADE] : [DBCE] = 9 : 16$.

6. Worked example 2 — rotunda: arc, sector, tangent, inscribed angle

Problem. A circular rotunda in Ermita has radius $21$ m. A landscaped sector is bounded by two radii meeting at a central angle of $120^\circ$. Using $\pi \approx \frac{22}{7}$: (a) how much edging borders the curved side? (b) what is the sector's area? (c) a straight walkway is tangent to the rotunda at $T$, and a lamp post $P$ on that walkway is $28$ m from $T$ — how far is $P$ from the centre $O$? (d) a statue $S$ stands on the major arc; what angle do the two sector endpoints subtend at $S$?

1. Fraction of the circle: $\frac{120^\circ}{360^\circ} = \frac{1}{3}$.
2. $L = \frac{1}{3}\cdot 2\pi(21) = 14\pi \approx 14\times\frac{22}{7} = 44$ m of edging.
3. $A_{\text{sec}} = \frac{1}{3}\pi(21)^2 = 147\pi \approx 147\times\frac{22}{7} = 462$ m$^2$.
4. Tangent–radius gives right angle $\angle OTP$, so $OP = \sqrt{21^2+28^2} = \sqrt{441+784} = \sqrt{1225} = 35$ m — this is the $3$-$4$-$5$ triple scaled by $7$.
5. The inscribed angle at $S$ intercepts the $120^\circ$ minor arc, so it measures $\frac{1}{2}(120^\circ) = 60^\circ$.

7. Common traps

• Co-interior angles are supplementary, not congruent — the most reversed rule in the whole topic.
• $360^\circ$ is the total of the exterior angles; dividing by $n$ gives one exterior angle only when the polygon is regular.
• The triangle inequality is strict: $5, 7, 12$ collapses into a straight segment, not a triangle.
• SSA and AAA prove nothing about congruence.
• Arc length uses $2\pi r$; sector area uses $\pi r^2$. Mixing them is the standard circle distractor.
• Substituting the diameter where the formula wants $r$ quadruples an area.
• Trapezoid and parallelogram areas need the perpendicular height, never the slanted side.
• For similar figures the area ratio is $k^2$, not $k$.