AE; then DE is a rectangular parallelogram of the same altitude with, and equal to the triangle ABC (Eucl. i. 42.). The perimeter of ABC is greater than that of DE. Because ABAC, and BE = EC, the perimeter of ABC is double of AC and EC together; also the perimeter of DE is double of AE and EC together. But since AEC is a right angle, AC is greater than AE; and .. the perimeter of ABC greater than that of DE. (28.) If from one of the acute angles of a rightangled triangle, a line be drawn to the opposite side; the squares of that side and the line so drawn are together equal to the squares of the segment adjacent to the right angle and of the hypothenuse. Let ABC be a right-angled triangle, and from A let AD be drawn to the opposite side; the squares of AD and BC are together equal to the squares of AC and BD. Ꭰ B For the squares of AD and BC together are equal to the squares of AB, BD and BC, i. e. to the squares of AC and BD; since the squares of AB and BC are equal to the square AC. (29.) In any triangle if a line be drawn from the vertex at right angles to the base, the difference of the squares of the sides is equal to the difference of the squares of the segments of the base. S From A the vertex of the triangle ABC, let AD be drawn perpendicular to the base; the difference of the squares of AB, AC is equal to the difference of the squares of BD, DC. For since ABD is a right-angled triangle, the square of AB is equal to the squares of AD, BD; and since ADC is a right-angled triangle, the square of AC is equal to the squares of AD, DC; whence the difference of the squares of AC and AB is equal to the difference of the squares of CD and DB. (30.) In any triangle, if a line be drawn from the vertex bisecting the base; the sum of the squares of the two sides of the triangle is double the sum of the squares of the bisecting line and of half the base. From the vertex A of the triangle DE From A draw AE perpendicular to BC; Then (Eucl. ii. 12.) AB2=AD2+DB2+2 BD × DE, and (Eucl. ii. 13.) AC2=AD2+DC2 - 2 CD × DE = AD2 + DB'— 2 BD × DE, whence AB+ AC2 = 2 AD2+2 DB2. (31.) If from the three angles of a triangle lines be drawn to the points of bisection of the opposite sides; the squares of the distances between the angles and the common intersection are together one third of the of the sides of the triangle. From the angles of the triangle ABC, let lines be drawn to the middle points of the opposite sides, intersecting each other in G; the sum of the squares of AG, GB, GC is one third of the sum of the squares of AB, BC, CA. squares B D E Join EF. Then AB2 + AC2 = 2 AE2 + 2 EB2, AB2 + BC2 = 2 AF2 + 2 FB*, AC2 + BC2 = 2 AD2 + 2DC3, H F .. AB2 + BC2 + CA2 = AE2 + BF2 + CD2+AF2+ [EB2 + AD2, Now the sum of the squares of AF, EB, AD is equal to one fourth of the sum of the squares of AB, BC, CA; whence three fourths of the sum of the squares of AB, BC, CA will be equal to the sum of the squares of AE, BF, CD; or three times the sum of the squares of AB, BC, CA is equal to four times the sum of the squares of AE, BF, CD. Now BG: GF:: BA: EF:: BC: CE :: 2 : 1, .. BG: BF :: 2 : 3, and BG2 : BF2 :: 4 : 9, whence 4 BF 9 BG2. And the same being true of each of the rest, three times the sum of the squares of AB, BC, CA, is equal to nine times the sum of the squares of AG, BG, CG; .. the sum of the the sum of the squares of AB, BC, CA is three times the sum of the squares of AG, BG, CG. COR. If from the angles of a triangle lines be drawn to the points of bisection of the opposite sides, the squares of those lines together are to the squares of the sides of the triangle as 3: 4. (32.) If from any point within or without any rectilineal figure, perpendiculars be let fall on every side ; the sum of the squares of the alternate segments made by them will be equal. BF G E Let ABCD be any quadrilateral figure (the demonstration being the same whatever be the number of sides). From any point I let perpendiculars IE, IF, IG, IH be drawn; AE2+ BF+GC2+DH2 = EB2 + FC2 + GD2+AH2. From I draw lines to each of the angles; = H then AE + EI2 (Al' = ) AH2 + HI3, whence, AE+BF2+CG2 + DH2 = EB2 + FC2 + GD2+HẨ. (33.) If from any point within a rectangular parallelogram lines be drawn to the angular points; the sums of the squares of those which are drawn to the opposite angles are equal. Let ABCD be a rectangular parallelogram, and F any point within it; join FA, FB, FC, FD; the squares of FA and FC are together equal to the squares of FB and FD. A B D Draw the diagonals AC, BD; and join FE. Because the triangles ADC, BDC are similar and equal, AC= BD; and .. their halves, AE and DE, are equal. Now (iv. 30.) FD1+FB2 = 2 DE2 + 2 EF2, =2AE+2 EF = AF2+ FC2. (34.) The squares of the diagonals of a parallelogram are together equal to the squares of the four sides. Let ABCD be a parallelogram, whose diagonals are AC, BD; the squares of AC, BD are together equal to the squares of AB, BC, CD, DA. Since DB is bisected by AC, D 2 AE2 + 2 ED2 = AD + AB', and for the same reason, 2 CE2 + 2 ED2 = CD2 + CB, A E .. 4AE2+4 ED' = AD' + AB2 + CB2+CD', (35.) If two sides of a trapezium be parallel to each other; the squares of its diagonals are together equal to the squares of its two sides which are not parallel and twice the rectangle contained by its parallel sides. A Let the sides AB, DC of the trapezium ABCD be parallel; draw the diagonals AC, BD; the squares of AC and BD, are together equal to the squares of AD and BC, and twice the rectangle AB, DC. Let fall the perpendiculars CE, DF. B Then (Eucl. ii. 12.), DB2= DA2 + AB2+2AB × AF, and AC CB + AB +2 AB × BE, = |