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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Full neighborhood
Depends on (10)
- I.5Proposition I.5In isosceles triangles the angles at the base are equal to one another; and if the equal straight lines be produced…
- I.10Proposition I.10To bisect a given finite straight line.
- I.11Proposition I.11To draw a straight line at right angles to a given straight line from a given point on it.
- I.29Proposition I.29A straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle…
- I.31Proposition I.31Through a given point to draw a straight line parallel to a given straight line.
- I.32Proposition I.32In any triangle, if one of the sides be produced, the exterior angle is equal to the two interior and opposite angles,…
- I.34Proposition I.34In parallelogrammic areas the opposite sides and angles are equal to one another, and the diameter bisects the areas.
- I.47Proposition I.47In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides…
- 2Common notion 2If equals be added to equals, the wholes are equal.
- 3Common notion 3If equals be subtracted from equals, the remainders are equal.
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