Proposition·Untested·2605.00009

Proposition II.9

If a straight line be cut into equal and unequal segments, the squares on the unequal segments of the whole are double of the square on the half and of the square on the straight line between the points of section.

Proof

Let be bisected at (I.10) and cut unequally at . At draw at right angles to (I.11), with . Join and ; through draw parallel to (I.31), meeting at ; through draw parallel to (I.31), meeting at . Join . Since is a right angle and , the triangle is right-isoceles, and half a right angle (I.5; I.32). Similarly half a right angle. Hence is a right angle (Common Notion 2), and triangle is right-angled at . By I.47: \[ AB^2 \;=\; AE^2 + EB^2. \] Now (I.47 in , plus ). Similarly inside the right triangles formed by the perpendicular at on , I.47 plus the fact that (which one shows from the parallel construction) yields, after Common Notions 2 and 3: \[ AD^2 + DB^2 \;=\; 2\cdot AC^2 + 2\cdot CD^2. \]

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