ABC is equal to AED, the remainder CG is thus equal to the remainder GD. And if we join AD then the angles at L are inferred (to be) right-angles, and CD (is inferred to be) double CL [Prop. 1.4]. So, for the same (reasons), the (angles) at M are also right-angles, and AC (is) double CM. Therefore, since angle ALC (is) equal to AMF, and (angle) LAC (is) common to the two triangles ACL and AMF, the remaining (angle) ACL is thus equal to ΖΑ. the remaining (angle) MFA [Prop. 1.32].