The areas of two triangles which have 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 D A' Hyp. In triangles ABC and A'B'C', To prove AABC A A'B'C' A'B' x A'C ' Proof. Draw... Elements of Geometry - Page 146by George Albert Wentworth - 1881 - 250 pagesFull view - About this book
| Daniel Cresswell - Euclid's Elements - 1817 - 454 pages
...has to the aggregate of the two chords that are next to it. PROP. VI. (XVII.) If two trapeziums have **an angle of the one equal to an angle of the other, and** if, also, the sides of the two figures, about each of their angles, be proportionals, the remaining... | |
| Adrien Marie Legendre - Geometry - 1819 - 574 pages
...proportional to the sides FG, GH, so that AB:FG::BC:GH. It follows from this, that the triangles ABC, FGH, **having an angle of the one equal to an angle of the other and the** sides about the equal angles proportional, are similar (208), consequently the angle ECA — GHF. These... | |
| Daniel Cresswell - Geometry - 1819 - 446 pages
...:HE::AF:AE; that is, FG is to GE in the given ratio. PROP. XVII. 23. THEOREM. If two trapeziums have **an angle of the one equal to an angle of the other, and** if, also, the sides of the two figvres, about each of tJieir angles, be proportionals, the remaining... | |
| Peter Nicholson - Architecture - 1823 - 210 pages
...to the sum of the two lines AD, DB, therefore AB2 = AC2 THEOREM 63. 161. Two triangles, which have **an angle of the one equal to an angle of the other,** are to each other as the rectangle of the sides about the equal Suppose* the two triangles joined,... | |
| Adrien Marie Legendre, John Farrar - Geometry - 1825 - 280 pages
...putting CM in the place of CA we shall have CP : CM : : CM : CQ ; consequently the triangles CPM, CQM, **having an angle of the one equal to an angle of the other and the** sides about the equal angles proportional, are similar (203) ; therefore MP : MQ : : CP : CM or CA.... | |
| Adrien Marie Legendre - Geometry - 1825 - 276 pages
...putting CM in the place of CA we shall have CP : CM : : CM : CQ ; consequently the triangles CPM, CQM, **having an angle of the one equal to an angle of the other and the** sides about the equal angles proportional, are similar (20") ; therefore MP:MQ::CP : CM or CA. But... | |
| Adrien Marie Legendre - 1825 - 570 pages
...properties of triangles involve those of all figures. THEOREM. 208. Two triangles, which have an ungle **of the one equal to an angle of the other and the** sides about these angles proportional, are similar. Demonstration. Let the angle A = D (fig. 122),... | |
| Walter Henry Burton - Astronomy - 1828 - 84 pages
...F, are equal; and so, if 'the angles at F had been supposed equal, the triangles would have had each **angle of the one equal to an angle of the other, and the** side CF lying between correspondent angles in each; whence also DF is equal to FE. Is this sufficiently... | |
| George Darley - Geometry - 1828 - 190 pages
...equal." Here we have a criterion whereby to judge of the equality of two triangular surfaces, which have **an angle of the one equal to an angle of the other.** For example : ABCD is a road cutting off a triangular field AOB. It is desirable that the line of road... | |
| Timothy Walker - Geometry - 1829 - 138 pages
...vertices by the space of a quadrant, the sides will become parallel each to each. 3. — When they have **an angle of the one equal to an angle of the other, and the** sides including these angles proportional — . Thus if the F45 angle A=A (fig. 45), and if AB : AD... | |
| |