338. In any triangle, if a medial line be drawn from the vertex to the base: I. The sum of the squares on the two sides is equivalent to twice the square on half the base, increased by twice the square on the medial line; II. The difference of the squares on the two sides is equivalent to twice the product of the base by the projection of the medial line upon the base. In the triangle ABC let AM be the medial line and MD the projection of A M upon the base B C. Also let A B be greater than A C. Since ABA C, the ZA MB will be obtuse and the ZAMC will be acute. Then AB2 = BM2 + A M2 + 2 B MX MD, § 336 (in any obtuse ▲ the square on the side opposite the obtuse is equivalent to the sum of the squares on the other two sides increased by twice the product of one of those sides and the projection of the other on that side); (in any ▲ the square on the side opposite an acute is equivalent to the sum of the squares on the other two sides, diminished by twice the product of one of those sides and the projection of the other upon that side). Add these two equalities, and observe that B M = MC. Subtract the second equality from the first. Then A B2 — A C2 = 2 BC X MD. Q. E. D. PROPOSITION XII. THEOREM. 339. The sum of the squares on the four sides of any quadrilateral is equivalent to the sum of the squares on the diagonals together with four times the square of the line joining the middle points of the diagonals. In the quadrilateral ▲ BCD, let the diagonals be A C and B D, and FE the line joining the middle points of the diagonals. We are to prove AB2+ BC2 + C D2 + DA2 = A C2 + BD2 + 4 E F2. (the sum of the squares on the two sides of a ▲ is equivalent to twice the square on half the base increased by twice the square on the medial line to the base), (the sum of the squares on the two sides of a ▲ is equivalent to twice the square on half the base increased by twice the square on the medial line to the base). Substitute in the above equality for (BE + DE2) its equivalent; 2 then AB2 + BC2 + CD2 + DÃ2 = 4 (40)2 + 4 (BD)2 = 2 +4 EF2 A C2 + BD2 + 4 E F2 Q. E. D. 340. COROLLARY. The sum of the squares on the four sides of a parallelogram is equivalent to the sum of the squares on the diagonals. PROPOSITION XIII. THEOREM. 341. Two triangles having an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. (A having the same altitude are to each other as their bases). (A having the same altitude are to each other as their bases). § 326 § 326 Q. E. D. PROPOSITION XIV. THEOREM. 342. Similar triangles are to each other as the squares on their homologous sides. A B A' B Let the two triangles be AC B and A'C' B'. (two are to each other as the products of their bases by their altitudes). (the homologous altitudes of similar ▲ have the same ratio as their homolo gous bases). PROPOSITION XV. THEOREM. 343. Two similar polygons are to each other as the squares on any two homologous sides. Let the two similar polygons be A B C, etc., and From the homologous vertices A and A' draw diagonals. (similar polygons have their homologous sides proportional); The AABC, ACD, etc., are respectively similar to A'B'C', A'C' D', etc., $294 (two similar polygons are composed of the same number of ▲ similar to each other and similarly placed). (similar are to each other as the squares on their homologous sides), |