# Proposition 31

In right-angled triangles the figure on the side opposite the right angle equals the sum of the similar and similarly described figures on the sides containing the right angle.

Let ABC be a right-angled triangle having the angle BAC right.

I say that the figure on BC equals the sum of the similar and similarly described figures on BA and AC.

I.12 VI.8

Then, since in the right-angled triangle ABC, AD has been drawn from the right angle at A perpendicular to the base BC, therefore the triangles DBA and DAC adjoining the perpendicular are similar both to the whole ABC and to one another.

VI.Def.1

And, since ABC is similar to DBA, therefore BC is to BA as BA is to BD.

VI.19,Cor

And, since three straight lines are proportional, the first is to the third as the figure on the first is to the similar and similarly described figure on the second.

Therefore BC is to BD as the figure on BC is to the similar and similarly described figure on BA.

V.24

For the same reason also, BC is to CD as the figure on BC is to that on CA, so that, in addition, BC is to the sum of BD and DC as the figure on BC is to the sum of the similar and similarly described figures on BA and AC.

But BC equals the sum of BD and DC, therefore the figure on BC equals the sum of the similar and similarly described figures on BA and AC.

Therefore, in right-angled triangles the figure on the side opposite the right angle equals the sum of the similar and similarly described figures on the sides containing the right angle.

Q.E.D.

## Guide

This proposition is a generalization of I.47 where the squares in I.47 are replaced by any similar rectilinear figures.

 Draw a square ABCD with diameters AC and BD meeting at E. Circumscribe a semicircle AGBHC about the right isosceles triangle ABC. Draw the arc AFC from A to C of a circle with center D and radius DA. Hippocrates finds the area of the lune formed between the semicircle AGBHC and the arc AF as follows. Note that there are three segments of circles in the diagram; two of them are small, namely, AGB with base AB, and BHC with base BC, and one is large, namely, AFC with base AC. The first two are congruent, and the third is similar to them since all three are segments in quarters of circles. 