DCF, are equiangular; also (constr. and S. 19. 3. cor. 1.) FB=FC; .. (E. 4. 1.) KB: BD:: CF or BF:CD; .. (E. 16. 5.) KB: BF :: BD: CD; .. (constr. and S. 4. 5.) CD is the required part of BD. PROP. X. 16. PROBLEM. From a given triangle to cut off a rhombus; the base of the rhombus being part of the base of the triangle, and having its extremity in a given point of that base. Let ABC be the given A, and D the given point in its base BC: It is required to cut off from the AABC a rhombus, having its base in BC, and terminated by the given point D. .::Draw AD, and produce it; from the centre B, at the distance BC, describe a circle, cutting AD produced in E, and join B, E; .. BE=BC; through D draw (E. 31. 1.) DF parallel to EB; also through F draw FG parallel to BC, and through G draw GH parallel to FD or BE; .. the figure FDHG is a □: And since (constr. and E. 28. 1.) the A BAC, FAG, are equiangular, as are, also, the ABE, AFD, .. (E. 4. 6.) And (E. 11. 5.) AB: BC or BE:: AF: FG: AF: FG:: AF: FD; .. (E. 9. 5.) FGFD: But (E. 34. 1.) FG = DH, and FD = GH; ... the figure FDHG, having its base DH in BC, and terminated by the given point D, is a rhombus. PROP. XI. 17. THEOREM. If two triangles have one angle of the one, equal to one angle of the other, and also another angle of the one, together with another angle of the other, equal to two right angles, the sides about the two remaining angles shall be portionals. pro Let the two ABC, DEF, have the BAC= ▲ EDF, and another, as ACB, of the one A, together with another, as DEF, of the other, F A B E equal to two right angles: Then AB: BC:: DF : FE. From F draw (S. 25. 1.) FG making with DE an FGE=/ FEG; .. (E. 6. 1.) FG-FE: And since (hyp.) the ACB + 2 DEF = two right, and that (E. 13. 1.) the DGF+< FGE two right,.. the 2 ACB + DEF= <DGF+ FGE; but (constr.) the FGE= FEG; .. the ACBDGF; and (hyp.) the BAC=▲ GDF; ... (S. 26. 1.) the two ACB, DGF, are equiangular; ... (E. 4. 1.) AB: BC:: DF: FG or FE. PROP. XII. 18. THEOREM. If, from the extremities of the base of a given triangle, there be drawn two straight lines, both on the same side of the base, and each equal to the adjacent side, and making with that side an angle equal to the vertical angle of the triangle, then the straight lines which join the extremities of the lines so drawn, and the further extremities of the base, shall cut off, from the sides, equal segments towards the vertex; and each of those segments shall be a mean proportional between the other segments, that are towards the base. From the extremities B and C of the base BC, D B A M E of the A ABC, let BD be drawn (E. 31. 1.) parallel to AC, and made equal to AB; and let CE be drawn parallel to AB, and made equal to AC; so that (E. 29. 1.) each of the ABD, ACE, is equal to the vertical ▲ BAC; also, let DC and EB be drawn, cutting AB and AC in L and M, respectively: Then AL= AM; and BL:LA:: AM or LA: MC. For (constr. and E. 15. 1.) the are equiangular, as are, also, the DLB, ALC, EMC, AMB; .. (E. 4. 6.) DB or AB: AC:: BL: LA; Again (E. 4. 6.) CE or AC: AB:: CM: MA; .. (E. 18.5.) AC+AB: AB:: AC: AM; .. (E. 16. 5.) AC+ AB: AC::AB: AM; .. (E. 9. 5.) AL=AM. Also, since it has been shewn, that AB: AC:: BL: LA, and AC:AB:: CM: MA, (S. 2. 5.) AB: AC:: AM: CM; .. (E. 11. 5.) BL: LA :: AM or LA: MC. PROP. XIII. 19. THEOREM. If at the extremities of the hypote nuse of a right-angled triangle two straight lines be drawn, on the same side of the hypotenuse as the right angle, each equal to, and each perpendicular to, the adjacent side, the two straight lines joining each of their extremities and the further extremity of the hypotenuse, shall cut each other in the same point of the perpendicular drawn to the hypotenuse from the right angle. From the extremities A and B, of the hypo tenuse BC of the right-angled ▲ ABC, let BD |