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 and Trigonometry - Side 86av Adrien Marie Legendre - 1836 - 359 siderUten tilgangsbegrensning - Om denne boken
| George Albert Wentworth - 1881 - 250 sider
...squares on the diagonals. GEOMETRY. — BOOK IV. PROPOSITION XIII. THEOREM. 3-41. Two triangles having **an angle of the one equal to an angle of the other** are to each other an the products of the sides including the equal angles. Let the triangles ABC and... | |
| Great Britain. Education Department. Department of Science and Art - 1882
...the ratio of AN to NB is the duplicate of the ratio of AM to MB. 2. If two triangles of equal area **have an angle of the one equal to an angle of the other,** prove that the sides about the equal angles are reciprocally proportional. 3. Shew how to divide a... | |
| Mathematical association - 1883
...two adjoining sides of the one respectively equal to two adjoining sides of the other, and likewise **an angle of the one equal to an angle of the other;** the parallelograms are identically equal. [By Superposition.] COR. Two rectangles are equal, if two... | |
| Evan Wilhelm Evans - 1884 - 155 sider
...NAM equal to SB. Draw AO parallel to BC. ANC = ACN = CAO. ANC = CBA + BAN. Complete the proof. 24. **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 in- _ eluding the equal angles. See Theo. VII. BAC :... | |
| Mathematical association - 1884
...two adjoining: sides of the one respectively equal to two adjoining sides of the other, and likewise **an ang:le of the one equal to an angle of the other** ; the parallelograms are identically equal. Let ABCD, EFGH be two parallelograms having the angle ABC... | |
| Association for the improvement of geometrical teaching - 1884
...two adjoining sides of the one respectively equal to two adjoining' sides of the other, and likewise **an angle of the one equal to an angle of the other** ; the parallelograms are identically equal. Let ABCD, EFGH be two parallelograms having the angle ABC... | |
| William Kingdon Clifford - 1885 - 271 sider
...proposition about parallel lines.1 The first of these deductions will now show us that if two triangles **have an angle of the one equal to an angle of the other and the sides containing** these angles respectively equal, they must be equal in all particulars. For if we take up one of the... | |
| Lewis Carroll - 1885 - 275 sider
...have two adjacent sides of the one respectively equal to two adjacent sides of the other, and likewise **an angle of the one equal to an angle of the other** ; the Parallelograms are identically equal.' This might be a useful exercise to set ; but really it... | |
| W. E. BYERLY - 1887
...B'C' A'B" hence AD BC A'D' X B'C' and we have ABC A' B' C' EXERCISE. Theorem. — 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. Suggestion. Let ADE and... | |
| George Albert Wentworth - 1888 - 386 sider
...other angular points of the polygon. D AREAS OF POLYGONS. PROPOSITION VII. THEOREM. 374. 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. Let the triangles ABC and... | |
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