SAT · Geometry and Trigonometry · TOPIC LESSON
Turn angle facts, matching vertices, and scale factors into reliable geometric equations.
BY THE END, YOU CAN
BEFORE YOU START
Original educational lesson. Work through the examples before revealing the practice answers. This lesson supports SAT Math and Reading and Writing skill study; it does not reproduce official questions or guarantee a score.
LEARN THROUGH ORIGINAL PRACTICE
The teaching passages, worked examples and exercises in this lesson are original. They are not official College Board questions or past-exam reproductions. Blueprint and observed-practice analytics describe separate factual metadata; those source questions are not displayed here.
01 / UNDERSTAND & APPLY
A geometric diagram is a model of stated information. Right-angle squares, matching tick marks, parallel-arrow marks, and explicit angle labels carry meaning. An angle that merely looks like 90° is not established as a right angle. A triangle that appears symmetric is not necessarily isosceles.
In ∠ABC, the middle letter B is the vertex; its rays pass through A and C. A segment has two endpoints, a ray extends from one endpoint in one direction, and a line extends both ways. Distinguishing these objects matters when an angle is described as exterior or a side is extended.
Keep a compact chain of reasons: given parallel lines → equal alternate interior angles → equal triangle angles → similarity. This prevents importing an attractive but unsupported relationship into an equation.
ORIGINAL PRACTICE · 1
No
Visual appearance is insufficient. Equality must be given or follow from a theorem, such as vertical angles or corresponding angles formed by parallel lines.
02 / UNDERSTAND & APPLY
Two intersecting lines create two pairs of opposite, or vertical, angles. Each vertical pair has equal measure. Adjacent angles whose noncommon rays form a straight line make a linear pair and sum to 180°. These two relationships often determine all four angles from one.
Vertical angles are equal because each is supplementary to the same adjacent angle. This is a reason, not a memorized visual trick. Two adjacent angles do not have to be supplementary unless their outer rays form a straight line.
Write the correct relationship before solving algebra. If the expressions label vertical angles, set them equal. If they label a linear pair, add them to 180. After solving, substitute into both expressions and check that every angle is positive and less than 180° when it is an ordinary nonreflex angle.
ORIGINAL WORKED EXAMPLE
Two vertical angles measure (4x + 9)° and (7x − 24)°. Find x and the measure of an adjacent angle.
Answer x = 11; adjacent angle 127°
ORIGINAL PRACTICE · 1
103°
Their sum is 180: 5t + 25 = 180, so t = 31. The angles are 77° and 103°, and the requested larger one is 103°.
03 / UNDERSTAND & APPLY
A transversal intersects two lines at distinct points. If those lines are parallel, corresponding angles are equal, alternate interior angles are equal, and interior angles on the same side of the transversal are supplementary. The same-side interior relationship is a sum rule, not an equality rule.
Corresponding angles occupy the same relative position at the two intersections. Alternate interior angles lie between the parallel lines and on opposite sides of the transversal. Identify the positions before using a rule; letters or numbers next to angles do not imply a pairing.
The converses are useful too: equal corresponding angles or equal alternate interior angles establish that two coplanar lines cut by a transversal are parallel. Without parallelism given or established, these cross-intersection equalities cannot simply be assumed.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
ORIGINAL WORKED EXAMPLE
Parallel lines are cut by a transversal. Same-side interior angles have measures (3x + 12)° and (5x − 8)°. Find both angles.
Answer 78° and 102°
ORIGINAL PRACTICE · 1
∠3 = 60°, ∠4 = 120°, ∠5 = 60°, ∠6 = 120°
Angle 3 is vertical to angle 1. Angle 4 is a linear pair with angle 1. Angle 5 corresponds to angle 1, and angle 6 forms a linear pair with angle 5.
ORIGINAL PRACTICE · 2
No
Parallel lines require equal alternate interior angles. The unequal measures rule out parallelism; being drawn nearly parallel does not change that.
04 / UNDERSTAND & APPLY
The three interior angles of a triangle sum to 180°. A right triangle therefore has two acute angles summing to 90°. An exterior angle formed by extending one side equals the sum of the two nonadjacent interior angles, because it is supplementary to the adjacent interior angle.
Choose the shorter route. If two remote interior angles are known and an exterior angle is requested, add them directly. If an exterior angle and one remote angle are known, subtract to find the other. Include only the two remote angles in the exterior-angle sum.
A triangle’s side lengths must satisfy the triangle inequality: any two sum to more than the third. Equality creates a flat, degenerate configuration. Within a triangle the longest side lies opposite the largest angle; equal sides lie opposite equal angles.
A + B + C = 180°
The interior angle sum.
Exterior angle at C = A + B
The exterior ray extends a side through C; A and B are the remote interior angles.
ORIGINAL WORKED EXAMPLE
A triangle has an exterior angle of 128°. Its two remote interior angles measure (2x + 6)° and (3x + 12)°. Find its largest interior angle.
Answer 78°
ORIGINAL PRACTICE · 1
105°
The exterior angle equals the sum of the two remote angles: 37 + 68 = 105. Equivalently the third interior angle is 75°, whose supplement is 105°.
ORIGINAL PRACTICE · 2
5 < x < 17
The third side must exceed the difference 11 − 6 and be less than the sum 11 + 6. Endpoints are excluded because they would make a degenerate triangle.
05 / UNDERSTAND & APPLY
An isosceles triangle has at least two equal sides, and the angles opposite those sides are equal. Conversely, equal angles imply equal opposite sides. An equilateral triangle has three equal sides and three 60° angles.
Carefully identify the vertex angle between the two equal sides. The other two are the base angles. The base angles are each half of 180° minus the vertex angle; the vertex angle is 180° minus twice a base angle.
An altitude from the vertex of an isosceles triangle to its base also bisects the base and the vertex angle. This follows from the congruent right triangles it creates. An arbitrary triangle’s altitude does not generally bisect its base.
ORIGINAL WORKED EXAMPLE
Triangle ABC has AB = AC, and ∠A = 44°. Find ∠B and ∠C.
Answer ∠B = ∠C = 68°
ORIGINAL PRACTICE · 1
QR
The side opposite ∠P is QR; the side opposite ∠Q is PR. Equal angles have equal opposite sides. PQ is not determined by the given information.
06 / UNDERSTAND & APPLY
Congruent triangles have equal corresponding sides and angles. A rotation, reflection, or translation changes position but not measurements, so a flipped triangle can still be congruent. The order in a statement such as ΔABC ≅ ΔPQR maps A to P, B to Q, and C to R.
Sufficient criteria include SSS (three corresponding sides), SAS (two corresponding sides and their included angle), ASA (two angles and their included side), and AAS (two angles and a corresponding nonincluded side). For right triangles, equal hypotenuses and one pair of equal legs establish congruence.
AAA establishes similarity, not necessarily congruence: all equilateral triangles share the same angles but can have different sizes. SSA is not a general congruence rule, because two sides and a nonincluded angle can sometimes fit two different triangles. Make sure an SAS angle is between the two known sides.
ORIGINAL WORKED EXAMPLE
Triangles ABC and PQR satisfy AB = PQ = 5, AC = PR = 8, and ∠A = ∠P = 60°. If BC = 7, what is QR, and why?
Answer QR = 7 by SAS congruence
ORIGINAL PRACTICE · 1
Similar but not congruent
Equal corresponding angles establish similarity. Different perimeters show that their corresponding side lengths are not all equal, so congruence is impossible.
07 / UNDERSTAND & APPLY
Similar triangles have equal corresponding angles and proportional corresponding sides. AA is sufficient: if two angle pairs agree, the third pair agrees by the angle sum. SSS similarity requires all three corresponding side ratios to match. SAS similarity requires two side ratios to match and the included angles to agree.
Write the vertex correspondence before building a proportion. For ΔABC ∼ ΔDEF, the pairs are AB ↔ DE, BC ↔ EF, and AC ↔ DF. All ratios in an equation must point in the same direction, such as second triangle divided by first triangle.
Similarity allows a common scale factor, while congruence is the special case k = 1. A positive scale factor changes lengths and perimeters by k but leaves angles unchanged. Equal area alone does not establish similarity; figures of different shapes can enclose the same area.
ORIGINAL WORKED EXAMPLE
ΔABC ∼ ΔDEF. AB = 8, BC = 12, AC = 10, and DE = 14. Find EF and DF.
Answer EF = 21; DF = 17.5
ORIGINAL PRACTICE · 1
SSS similarity; scale factor 3/2
After matching shortest to shortest, middle to middle, and longest to longest, the ratios are 9/6 = 12/8 = 15/10 = 3/2. Three equal side ratios establish similarity.
ORIGINAL PRACTICE · 2
35
KL corresponds to ST, so 14/ST = 2/5. Cross-multiplying gives 2ST = 70 and ST = 35. The larger triangle’s corresponding side should indeed be longer.
08 / UNDERSTAND & APPLY
When a segment joins two sides of a triangle and is parallel to the third side, the small triangle at the shared vertex is similar to the whole triangle by AA. The scale factor uses a small whole side divided by the corresponding large whole side.
The common trap is comparing a small side with only the leftover segment of the larger side. If D lies on AB, the whole side is AB = AD + DB. You may derive ratios of the leftover segments, but first establish the whole-triangle correspondence.
In the diagram, D and E are midpoints, so the smaller triangle is half the size in each length. Its angles agree with those of the whole triangle, and DE is half BC. The example below uses a different division ratio to demonstrate the general method.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
ORIGINAL WORKED EXAMPLE
D lies on AB and E lies on AC, with DE parallel to BC. AD = 4, DB = 6, and AE = 6. Find EC.
Answer EC = 9
ORIGINAL PRACTICE · 1
12
The small-to-large scale factor is 3/7. Since DE corresponds to BC, DE = (3/7)(28) = 12. Angles do not scale by 3/7; they stay equal.
09 / UNDERSTAND & APPLY
Similar triangles with corresponding length ratio k have area ratio k², because both a base and its matching altitude scale by k. If the area ratio is given, take its positive square root to recover the length ratio. Perimeters still scale by k.
Coordinate differences supply exact lengths and directions. Horizontal and vertical segment lengths are absolute coordinate differences. For general segments, the distance formula is the Pythagorean theorem: √((x₂ − x₁)² + (y₂ − y₁)²).
For nonvertical lines, equal slopes establish parallelism, while slopes whose product is −1 establish perpendicularity. Treat vertical and horizontal lines separately: they are perpendicular even though a vertical slope is undefined. Coordinate checks should confirm a relationship rather than replace missing information with an estimate.
ORIGINAL WORKED EXAMPLE
Two similar triangles have areas 45 and 80. The smaller triangle has perimeter 27. Find the larger perimeter.
Answer 36
ORIGINAL PRACTICE · 1
Area 24; perimeter 24
The horizontal and vertical legs are 6 and 8, so they are perpendicular. Area is (1/2)(6)(8) = 24. The third side is √(6² + 8²) = 10, giving perimeter 6 + 8 + 10 = 24. The numerical equality does not make area and perimeter the same measure.
10 / UNDERSTAND & APPLY
An angle-chasing problem usually starts with one of three structures: intersecting lines, parallel lines, or a triangle. Mark the immediate consequences, then connect them. Solving every angle is often unnecessary; find only those that link the given information to the target.
A length problem usually needs equality from congruence, a ratio from similarity, or a right-triangle calculation. Decide which relationship is justified before choosing an equation. One equal angle pair alone does not establish triangle similarity, and two matching sides alone do not establish triangle congruence.
Use a calculator to solve arithmetic-heavy proportions if useful, but retain exact fractions until the last step. Verify scale direction, units, and angle bounds. A smaller similar triangle cannot have a longer corresponding side, and scaling a triangle cannot alter its angle measures.
ORIGINAL PRACTICE · 1
42° and 150 square units
Similarity keeps angles fixed. Area scales by (5/2)² = 25/4, so the new area is 24 × 25/4 = 150. Multiplying 42° by the length scale factor would be invalid.
KEEP THE BIG IDEAS
READ WITH CONTEXT
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