The sum of the perpendiculars dropped from any point within an equilateral triangle to the three sides is constant, and equal to the altitude. Plane Geometry - Page 71by George Albert Wentworth - 1899 - 256 pagesFull view - About this book
| Adrien Marie Legendre - Geometry - 1882 - 194 pages
...construct on it a square, as before. fi. — Show that the sum of the. three perpendiculars, drawn **from any point within an equilateral triangle to the three sides, is** equal U the altitude of the triangle. Let ABC be an equilateral triangle, and BD its altitude. From... | |
| Charles Davies, Adrien Marie Legendre - Geometry - 1885 - 538 pages
...respectively 16, 12, 8, 4, and 2 units in length. 5. Show that the sum of the three perpendiculars drawn **from any point within an equilateral triangle to the three sides is** equal to the altitude of the triangle. 6. Show that the sum of the squares of two lines, drawn from... | |
| George Albert Wentworth - Geometry - 1888 - 264 pages
...the two Js, BF the altitude upon AC. Draw PG ± to BF, and prove the A PSG and PBD equal. . . P 74. **The sum of the perpendiculars dropped from any point...three sides is constant, and equal to the altitude.** HINT. Draw through the point a line II to the base, and apply Ex. 73. 75. What is the locus of all... | |
| Edward Mann Langley, W. Seys Phillips - 1890 - 538 pages
...from any point in the base is constant (see Ex. 133). Show also that the sum of the perpendiculars **from any point within an equilateral triangle to the three sides is constant.** Ex. 681.— ABC, DBC are two triangles on the same base BC : the line joining the vertices A and D... | |
| Euclid - Geometry - 1892 - 460 pages
...difference of its distances from the equal sides is constant. 24. The sum of the perpendiculars drawn **from any point within an equilateral triangle to the three sides is** equal to the perpendicular drawn from any one of the angular points to the opposite side, and is therefore... | |
| George Albert Wentworth - Geometry - 1896 - 296 pages
...is common. .-. A PBD = A PEG. .: PD = BG. PE= GF. .: PD + PE= BG + GF= BF. (g 148) (g 180) Ex. 74. **The sum of the perpendiculars dropped from any point...three sides is constant, and equal to the altitude.** Let ABC be an equilateral A, BC the base, AD the altitude, P any point within the A, PE, PF, PG _!•... | |
| Joe Garner Estill - 1896 - 214 pages
...triangles have the same ratio as any two homologous sides. Prove. 5. The sum of the perpendiculars **from any point within an equilateral triangle to the three sides is** equal to the altitude of the triangle. Prove. Boston University, September, 1896. TIME 1 H. 80 M. [Candidates... | |
| George D. Pettee - Geometry, Plane - 1896 - 272 pages
...parallelogram divides the parallelogram into two equivalent figures. 294. The sum of the perpendiculars drawn **from any point within an equilateral triangle to the three sides is** equal to the altitude of the triangle. 295. The area of a triangle is equal to the product of the semiperimeter... | |
| Webster Wells - Geometry - 1898 - 264 pages
...(Ex. 13, p. 173.) (Represent the diagonals by 2 a; and 2 y.) A j 43. The sum of the perpendiculars **from any point within an equilateral triangle to the three sides is** equal to the altitude of the triangle. "/-'' DE 44. The longest sides of two similar polygons are 18... | |
| George Albert Wentworth - Geometry - 1899 - 498 pages
...(Why ?) .-. GF = PE. It remains to prove GB = PD. The rt. A PGB = the rt. A BDP. (Why ?) A — J- "B **Ex. 61. The sum of the perpendiculars dropped from...Then MD = PE. (Why ?) PG + PF = AM (Ex. 60). Ex. 62.** ABC and ABD are two triangles on the same base AB, and on the same side of it, the vertex of each triangle... | |
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