For, since the angle BAC is equal to the angle ADB,
for each is right,
and the angle at B is common to the two triangles ABC and ABD,
therefore the remaining angle ACB is equal to the remaining angle BAD; [I. 32]
therefore the triangle ABC is equiangular with the triangle ABD.
Therefore, as BC which subtends the right angle in the triangle ABC is to BA which subtends the right angle in the triangle ABD,
so is AB itself which subtends the angle at C in the triangle ABC to BD which subtends the equal angle BAD in the triangle ABD,
and so also is AC to AD which subtends the angle at B common to the two triangles. [VI. 4]
Therefore the triangle ABC is both equiangular to the triangle ABD and has the sides about the equal angles proportional.
Therefore the triangle ABC is similar to the triangle ABD. [VI. Def. 1]
Similarly we can prove that the triangle ABC is also similar to the triangle ADC;
therefore each of the triangles ABD, ADC is similar to the whole ABC.