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Overview

Euclidean geometry problems rely on angle chasing and similarity. Power of a point and cyclic quadrilaterals are frequent tools in AMC 12 and AIME.

Key Ideas

  • If a quadrilateral is cyclic, opposite angles sum to 180180^\circ.
  • Power of a point: for chords through a point, PAPB=PCPDPA\cdot PB=PC\cdot PD.
  • Similar triangles yield proportional sides and equal angles.

Worked Example

In a circle, chords ABAB and CDCD intersect at PP. If PA=3PA=3, PB=5PB=5, and PC=2PC=2, then PD=PAPBPC=152PD=\frac{PA\cdot PB}{PC}=\frac{15}{2}.

Practice Problems

StatusSourceProblem NameDifficultyTags
AMC 12Normal
Show TagsAngles, Circles
AIMEHard
Show TagsPower of a Point

Module Progress:

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