All 440 of the famous Geometry Puzzles by Catriona Agg made interactive with Live Geometry:

https://livegeometry.com/CatrionaAgg

Examples:
https://livegeometry.com/CatrionaAgg/three-regular-hexagons-and-a-semi-circle

I’ve been working through this standard geometry result and liked how one identity handles both the inequality and equality case.

Let a and b be adjacent side lengths of a parallelogram, with included angle θ, where 0° < θ < 180°.
Let c be the diagonal from that vertex to the opposite vertex.

Keep c and θ fixed.
The area is K = ab sin θ, and:
c² = (a − b)² + 2ab(1 + cos θ).

Using (1 + cos θ) tan(θ/2) = sin θ gives:
(c²/2) tan(θ/2) − K = ((a − b)²/2) tan(θ/2).

The right-hand side is nonnegative, so the maximum area is:
K_max = (c²/2) tan(θ/2).

Equality holds exactly when a = b: a rhombus.
At θ = 90°, this recovers the familiar square-maximises-area result for rectangles with a fixed diagonal.
The identity also measures exactly how far the area falls below the maximum.

Does anyone have a geometric interpretation of that deficit term?