Key Concepts
- 1BPT (Thales Theorem)
- 2AA Similarity
- 3SSS Similarity
- 4SAS Similarity
- 5Area ratio of similar triangles
- 6Pythagoras Theorem
- 7Converse of Pythagoras
- 8Converse of BPT
- 9BPT (Thales Theorem)
- 10AA Similarity
Important Formulas & Facts
DE ∥ BC in △ABC ⟹ AD/DB = AE/EC
If 2 angles of one △ = 2 angles of another △, they are similar.
If all 3 pairs of sides are proportional, the triangles are similar.
If one angle is equal and the including sides are proportional.
ar(△1)/ar(△2) = (side₁/side₂)²
In right △: Hypotenuse² = Base² + Perpendicular²
If AB² + BC² = AC², then ∠B = 90°
If a line divides two sides in the same ratio, it is parallel to the third side.
DE ∥ BC in △ABC ⟹ AD/DB = AE/EC
If 2 angles of one △ = 2 angles of another △, they are similar.
Must-Know Questions
Q1D and E are on AB and AC of △ABC with DE ∥ BC and AD:DB = 3:1. If EA = 3.3 cm, find AC.
AD/DB = AE/EC = 3/1. AE = 3.3, EC = 3.3/3 = 1.1. AC = 4.4 cm.
Q2Which similarity criterion requires two pairs of equal angles?
AA (Angle-Angle) criterion: If two angles of one triangle equal two angles of another, the triangles are similar.
Q3Assertion (A): In △ABC, DE ∥ BC with D on AB and E on AC. If AD/DB = 2/3, then AE/AC = 2/5. Reason (R): By BPT, AD/DB = AE/EC.
R: BPT gives AD/DB = AE/EC. True. A: AE/EC = 2/3. So AE/AC = AE/(AE+EC) = 2/(2+3) = 2/5. True. R explains A.
Q4Prove that if a line divides any two sides of a triangle in the same ratio, the line is parallel to the third side.
This is the converse of BPT. Given AD/DB = AE/EC. Suppose DE is not parallel to BC. Draw DE' ∥ BC. Then AD/DB = AE'/E'C (by BPT). So AE/EC = AE'/E'C. This means E and E' coincide. Hence DE ∥ BC.
Q5In △ABC, DE ∥ BC. If AD = x, DB = x-2, AE = x+2, EC = x-1, find x.
By BPT: AD/DB = AE/EC. x/(x-2) = (x+2)/(x-1). x(x-1) = (x+2)(x-2). x² - x = x² - 4. -x = -4. x = 4.
Practice Triangles
Reinforce what you just revised with practice questions