📌 Step 1: Apply the Triangle Proportionality Theorem
When a line is parallel to one side of a triangle and intersects the other two sides, it divides those sides proportionally.

📌 Step 2: Set up the proportion
AD/DB = AE/EC
6/4 = 9/EC

📌 Step 3: Cross-multiply and solve
6 × EC = 4 × 9
6 × EC = 36
EC = 36/6 = 6

💡 Tip: This theorem is also called the "Side Splitter Theorem." It works because DE ∥ BC creates similar triangles.

📌 Step 1: Recall the distance formula
d = √((x₂ − x₁)² + (y₂ − y₁)²)

📌 Step 2: Plug in the coordinates
A(2, 3) and B(6, 6):
d = √((6 − 2)² + (6 − 3)²)
d = √(4² + 3²)
d = √(16 + 9)

📌 Step 3: Solve
d = √25 = 5

💡 Tip: The distance formula is just the Pythagorean theorem applied to coordinates!

Using the Pythagorean theorem: h² + 9² = 15² → h² = 225 - 81 = 144 → h = 12 feet.

📌 Step 1: Identify the angle relationship
These are alternate interior angles (between the parallel lines, on opposite sides of the transversal).

📌 Step 2: Apply the theorem
Alternate interior angles are equal when lines are parallel:
3x + 10 = 5x − 30

📌 Step 3: Solve for x
10 + 30 = 5x − 3x
40 = 2x
x = 20

💡 Verification: 3(20) + 10 = 70° and 5(20) − 30 = 70°. ✓ They're equal!

First test whether R1's hypothesis is true. Substituting x = 19: m∠A = 2(19) + 5 = 43° and m∠B = 3(19) − 10 = 47°, so m∠A + m∠B = 90°. Because the hypothesis of R1 holds, the Law of Detachment gives the conclusion '△ABC is a right triangle.' That conclusion is exactly the hypothesis of R2, so a second application of the Law of Detachment (the Law of Syllogism links R1 and R2 into a single conditional) gives '△ABC is not equiangular.' This also agrees with the Triangle Sum Theorem, since an equiangular triangle has three 60° angles and cannot contain a 90° angle. The most tempting wrong choice is the 110° option: it comes from a sign slip, reading 3x − 10 as 3x + 10 and getting 57 + 10 = 67, then 43 + 67 = 110°, which would make R1's hypothesis false and stop the chain. Stopping at 'right triangle' is also incomplete — a valid conclusion of one rule must be pushed through any rule whose hypothesis it matches. Using the converse of R2 is invalid: a conditional and its converse are not logically equivalent.

📌 Step 1: Identify the right triangle
The ladder, wall, and ground form a right triangle where:
• The wall height = 12 ft (one leg)
• The ground distance = 5 ft (other leg)
• The ladder = hypotenuse (what we need)

📌 Step 2: Apply the Pythagorean Theorem
a² + b² = c²
12² + 5² = c²
144 + 25 = c²
169 = c²

📌 Step 3: Solve for c
c = √169 = 13 feet

💡 Tip: 5-12-13 is a common Pythagorean triple. Memorizing these saves time on the CBE!

Since M is the midpoint of AC: AM = CM.
Since M is the midpoint of BD: BM = DM.
∠AMB = ∠CMD (vertical angles are congruent).
Two sides and the included angle → SAS congruence theorem.

📌 Step 1: Identify the trig ratio
We know the hypotenuse (20) and want the side opposite to 35°.
Opposite and hypotenuse → use sine.

📌 Step 2: Set up the equation
sin(35°) = opposite / hypotenuse
0.574 = opposite / 20

📌 Step 3: Solve
opposite = 20 × 0.574 = 11.47 ≈ 11.5

💡 Remember SOH-CAH-TOA:
• Need opposite? Use sin (if you have hypotenuse) or tan (if you have adjacent)
• Need adjacent? Use cos (if you have hypotenuse) or tan (if you have opposite)

The walkway adds w to BOTH ends of each dimension, so the outer rectangle is (30 + 2w) by (20 + 2w). Using the area of a rectangle, (30 + 2w)(20 + 2w) = 936, so 600 + 100w + 4w² = 936, giving 4w² + 100w − 336 = 0, or w² + 25w − 84 = 0. The quadratic formula gives w = (−25 ± √(625 + 336))/2 = (−25 ± 31)/2, so w = 3 or w = −28. A width cannot be negative, so w = 3 ft; check: 36 × 26 = 936 ✓. The most tempting wrong value is 6 ft, which comes from writing (30 + w)(20 + w) = 936 — that adds the walkway to only one end of each side, forgetting that a border surrounds the pool on both sides. The value 3.36 comes from dividing the extra area 336 by the pool perimeter 100 and ignoring the four corner squares, and 28 is the rejected negative root taken as positive.