SAT Lines, Angles, and Triangles

This skill is about angle facts and triangle relationships: angles formed by intersecting lines, angles formed when a transversal cuts parallel lines, the angle sum of a triangle, isosceles triangles, and similar triangles, whose sides are proportional. Most questions combine one or two facts with a short algebra step.

• Angles that form a straight line add to $180^\circ$. Vertical angles (opposite each other where two lines cross) are equal.
• When a transversal crosses parallel lines, every acute angle is equal and every obtuse angle is equal, and any acute angle plus any obtuse angle is $180^\circ$.
• The angles of a triangle add to $180^\circ$. An exterior angle equals the sum of the two interior angles that are not next to it.
• In an isosceles triangle, the angles opposite the equal sides are equal.
• Two triangles are similar if two pairs of angles match (AA). Then all corresponding sides share one ratio.
• A segment drawn parallel to one side of a triangle cuts off a smaller triangle similar to the whole.

• Triangle: $\angle A + \angle B + \angle C = 180^\circ$.
• Exterior angle at $C$: $180^\circ - \angle C = \angle A + \angle B$.
• Similar triangles: $\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$, listed in matching order.
• Right triangle with altitude to the hypotenuse: the altitude $h$ splits the hypotenuse into $p$ and $q$ with $h^2 = pq$.

The angles of a triangle measure $(2x + 10)^\circ$, $(3x - 5)^\circ$, and $(x + 25)^\circ$. Find each angle.

1. Add and set equal to $180$: $6x + 30 = 180$, so $6x = 150$ and $x = 25$.
2. The angles are $2(25) + 10 = 60^\circ$, $3(25) - 5 = 70^\circ$, and $25 + 25 = 50^\circ$.

Check: $60 + 70 + 50 = 180$. Correct.

In triangle $PQR$, point $S$ is on $\overline{PQ}$ and point $T$ is on $\overline{PR}$, with $\overline{ST}$ parallel to $\overline{QR}$. If $PS = 6$, $SQ = 4$, and $ST = 9$, what is $QR$?

1. Because $\overline{ST} \parallel \overline{QR}$, the corresponding angles match, so triangle $PST$ is similar to triangle $PQR$.
2. Match the sides from the shared vertex $P$: $PS$ corresponds to the whole side $PQ = 6 + 4 = 10$.
3. Set up the ratio: $\frac{ST}{QR} = \frac{PS}{PQ}$, so $\frac{9}{QR} = \frac{6}{10}$.
4. Cross-multiply: $6 \cdot QR = 90$, so $QR = 15$.

Check: the scale factor from the small triangle to the large one is $\frac{10}{6} = \frac{5}{3}$, and $9 \cdot \frac{5}{3} = 15$. Correct.

• Using the piece instead of the whole side. In example 2, comparing $PS$ to $SQ$ gives $\frac{9}{QR} = \frac{6}{4}$ and the wrong answer $6$.
• Mismatching corresponding sides. Write the similarity in vertex order ($\triangle PST \sim \triangle PQR$) and read sides in the same order.
• Assuming same-side interior angles are equal. They add to $180^\circ$; only alternate and corresponding angles are equal.
• Picking the wrong equal angles in an isosceles triangle. The equal angles are opposite the equal sides. The angle between the equal sides is the different one.
• Answering $x$ instead of the angle. After solving for $x$, substitute back if the question asks for an angle measure.