to DF and CH: But AH is the rectangle contained by AD, DB, for DH is equal to DB; and DF together with CH is the gnomon CMG; therefore the rectangle AD, DB is equal to the gnomon CMG: To each of these add LG, which is equal to the square of CD (2. 4. Cor.); therefore the rectangle AD, DB, together with the square of CD, is equal to the gnomon CMG, together with LG: But the gnomon CMG and LG make up the whole figure CEFB, which is the square of CB; therefore the rectangle AD, DB, together with the square of CD, is equal to the square of CB. Wherefore, If a straight line, &c. Q. E. D. COR. From this proposition it is manifest, that the difference of the squares of two unequal lines AC, CD, is equal to the rectangle contained by their sum and difference. PROP. VI. THEOR. If a straight line be bisected and produced to any point, the rectangle contained by the whole line thus produced and the part of it produced, together with the square of half the line bisected, is equal to the square of the straight line which is made up of the half and the part produced. A CBD Let the straight line AB be bisected in C, and produced to D: the rectangle AD, DB, together with the square of CB, shall be equal to the square of CD. Upon CD describe the square CEFD, join DE, and through B draw BHG parallel to CE or DF, and through H draw KLM parallel to AD or EF, and through A draw AK parallel to CL or DM: Then, because AC is equal to CB, the rectangle AL is equal to the rectangle CH (1.36): But CH is equal to HF; therefore also AL is equal to HF: To K L H M GF each of these add CM; therefore the whole AM is equal to the gnomon CMG: But AM is the rectangle contained by AD, DB, for DM is equal to DB; therefore the rectangle AD, DB is equal to the gnomon CMG: To each of these add LG, which is equal to the square of CB; therefore the rectangle AD, DB, together with the square of CB, is equal to the gnomon CMG, and the figure LG: But the gnomon CMG and LG make up the whole figure CEFD, which is the square of CD; therefore the rectangle AD, DB, together with the square of CB, is equal to the square of CD. Wherefore, If a straight line &c. Q. E. D. PROP. VII. THEOR. If a straight line be divided into any two parts, the squares of the whole line and of one of the parts are equal to twice the rectangle contained by the whole and that part, together with the square of the other part. Let the straight line AB be divided into any two parts at the point C: the squares of AB, BC shall be equal to twice the rectangle AB, BC, together with the square of AC. A C B Upon AB describe the square ADEB, and construct the figure as in the preceding propositions : Then, because AG is equal to GE, to each of these add CK; therefore the whole AK is equal to the whole CE, and AK, CE are together double of AK: But AK, CE are the gnomon AKF, together with the square CK; therefore the gnomon AKF, together with the square CK, is double of AK: But H D K twice the rectangle AB, BC is double of AK, for BK is equal to BC; therefore the gnomon AKF, together with the square CK, is equal to twice the rectangle AB, BC: To each of these equals add HF, which is equal to the square of AC; therefore the gnomon AKF, together with the squares CK, HF, is equal to twice the rectangle AB, BC, together with the square of AC: But the gnomon AKF, together with the squares CK, HF, make up the whole figure ADEB and CK, which are the squares of AB and BC; therefore the squares of AB, BC, are equal to twice the rectangle AB, BC, together with the square of AC. PROP. VIII. PROP. IX. THEOR. If a straight line be divided into two equal and also into two unequal parts, the squares of the two unequal parts are together double of the square of half the line and of the square of the line between the points of section. Let the straight line AB be divided at the point C into two equal, and at D into two un equal parts: the squares of AD, DB shall be together double of the squares of AC, CD. From the point C draw CE at right angles to AB, and make CE equal to CA or CB, and join EA, EB; through D draw DF parallel to CE, and through F draw FG parallel to AB; and join AF: Then, because CA is equal to CE, the angle CAE is equal to the angle CEA; but the angle ACE is a right angle, therefore the two angles CAE, CEA are together equal to a right angle (1. 32); and they are equal to one another; therefore each of them is half a right angle: For the like reason each of the angles CBE, CEB is half a right angle: Therefore the whole angle AEB is a right angle: And because the angle GEF is half a right angle, and EGF a right angle, for it is equal to the interior and opposite angle ECB (1. 29), the remaining angle GFE is half a right angle; therefore the angle GEF is equal to the angle GFE, and the side GE to the side GF: Again, because the angle at B is half a right angle, and FDB a right angle, (for it is equal to the interior and opposite angle ECB,) therefore the remaining angle DFB is half a right angle; therefore the angle DBF is equal to the angle DFB, and the side DB to the side DF. A E G F CD B Now, because AC is equal to CE, the square of AC is equal to the square of CE, and therefore the squares of AC, CE are double of the square of AC: But the square of AE is equal to the squares of AC, CE, because ACE is a right angle (1. 47); therefore the square of AE is double of the square of AC: Again, because EG is equal to GF, the square of EG is equal to the square of GF, and therefore the squares of EG, GF are double of the square of GF: But the square of EF is equal to the squares of EG, GF; therefore the square of EF is double of the square of GF, that is, of the square of CD: And the square of AE is also double of the square of AC; therefore the squares of AE, EF are double of the squares of AC, CD: And the square of AF is equal to the squares of AE, EF, because AEF is a right angle; therefore the square of AF is double of the squares of AC, CD: And the square of AF is equal to the squares of AD, DF, because ADF is a right angle; therefore the squares of AD, DF are double of the squares of AC, CD: And DF is equal to DB; therefore the squares of AD, DB are together double of the squares of AC, CD. Wherefore, If a straight line &c. Q. E. D. PROP. X. THEOR. If a straight line be bisected and produced to any point, the squares of the whole line thus produced and of the part of it produced are together double of the square of half the line bisected, and of the square of the line made up of the half and the part produced. Let the straight line AB be bisected in C, and produced to D: the squares of AD, DB shall be together double of the squares of AC, CD. From the point C draw CE at right angles to AB and make it equal to CA or CB, and join AE, EB; through E draw E F B ען EF parallel to AB, and through D draw DF parallel to CE: And because the straight line EF meets the parallels EC, FD, the angles CEF, EFD are together equal to two right angles, and therefore the angles BEF, EFD are together less than two right angles: But If a straight line meets two straight lines, so as to make the two interior angles on the same side of it together less than two right angles, these straight lines, being continually produced, shall at length meet, upon that side on which are the angles which are less than two right angles; therefore EB, FD shall meet, if produced, towards B, D; let them meet in G, and join AG. Then, because CA is equal to CE, the angle CAE is equal to the angle CEA; but the angle ACE is a right angle; therefore each of the angles CAE, CEA is half a right angle: For the like reason, each of the angles CBE, CEB is half a right angle: There fore AEB is a right angle: And because the angle DBG is half a right angle (for it is equal to the |