Triangle Proportionality
If DE is parallel to side BC and hits the other two sides, it splits those sides in the same ratio: AD/DB = AE/EC. You can also write AD/AB = AE/AC. Keep the pieces matching: small-to-rest with small-to-rest, or small-to-whole with small-to-whole. Do not mix them.
Why: DE ∥ BC makes corresponding angles match, so ΔADE ~ ΔABC by AA (or SAS similarity on the shared angle). Matching sides then give the proportion. Corresponding angles also force the parallel in the first place if you already had SAS similarity.
Converse: if a line splits two sides proportionally, it is parallel to the third side.
Altitude on the hypotenuse
Right triangle ABC, right angle at B, altitude from B to hypotenuse AC at D. Then ΔABC ~ ΔADB ~ ΔBDC. The course calls this the Pieces of Right Triangles Similarity Theorem.
Each small triangle shares a 90° and one of the original acute angles, so AA with the big triangle and with each other.
From those similarities you can prove a² + b² = c². That is the Pythagorean Theorem, proved with similar triangles, not as a slogan. The converse: if a² + b² = c² on the longest side c, then the angle opposite c is 90°. The course's converse proof builds a right triangle with those legs and uses SSS plus CPCTC to copy the right angle.
Two proof shapes that keep showing up
Parallelogram with a diagonal: opposite sides parallel, so alternate interior angles match, the diagonal is Reflexive, ASA, the two triangles are congruent.
Two triangles sharing a vertex with a perpendicular through it: both get a 90°, plus a given matching pair, so AA similarity.
Crossed segments between parallels: vertical angles at the crossing plus alternate interior (or corresponding) from the parallels, AA.