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Summary
| Description |
English: Proof without words of Varignon's theorem by CMG Lee.
1. An arbitrary quadrilateral and its diagonals. 2. Bases of similar triangles are parallel to the blue diagonal. 3. Ditto for the red diagonal. 4. The base pairs form a parallelogram with half the area of the quadrilateral, Aq, as the sum of the areas of the four large triangles, Al is 2 Aq (each of the two pairs reconstructs the quadrilateral) while that of the small triangles, As is Al/4 (half linear dimensions yields quarter area) = Aq/2, and the area of the parallelogram is Aq − As = Aq − Aq/2 = Aq/2. |
| Source | Own work |
| Author | Cmglee |
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| Date/Time | Thumbnail | Dimensions | User | Comment | |
|---|---|---|---|---|---|
| current | 20:23, 24 September 2020 | | 512 × 683 (4 KB) | Cmglee | {{Information |description ={{en|1=Proof without words of Varignon's theorem by CMG Lee. 1. An arbitrary quadrilateral and its diagonals. 2. Bases of similar triangles are parallel to the blue diagonal. 3. Ditto for the red diagonal. 4. The base pairs form a parallelogram with half the area of the quadrilateral, ''A<sub>q</sub>'', as the sum of the areas of the four large triangles, ''A<sub>l</sub>'' is 2 ''A<sub>q</sub>'' (each of the two pairs reconstructs the quadrilateral) while tha... |
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