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718:{{Information |Description=Visual proof of Lee Sallows's triangle theorem by CMG Lee. 1. Lines from midpoints of edges of length ''a'', ''b'' and ''c'' (circles) to the opposite vertex split the triangle into 6 regions. 2. Each region is rotated 90 degrees outwards about its hinge, yielding 3 congruent triangles with sides equal and perpendicular to the line segments between the original triangle's vertices and its centroid (2''x'', 2''y'' and 2''z''). 3. The process can be repeated on each s... 43: 399: 311: 226: 231: 558:"}},"text\/plain":{"en":{"P275":"Creative Commons Attribution-ShareAlike 4.0 International"}}},"{\"value\":{\"entity-type\":\"item\",\"numeric-id\":50829104,\"id\":\"Q50829104\"},\"type\":\"wikibase-entityid\"}":{"text\/html":{"en":{"P275":" 95: 411:"}},"text\/plain":{"en":{"P2093":"Cmglee","P4174":"Cmglee"}}},"{\"value\":{\"entity-type\":\"property\",\"numeric-id\":4174,\"id\":\"P4174\"},\"type\":\"wikibase-entityid\"}":{"text\/html":{"en":{"":" 554:"}},"text\/plain":{"en":{"":"copyright license"}}},"{\"value\":{\"entity-type\":\"item\",\"numeric-id\":18199165,\"id\":\"Q18199165\"},\"type\":\"wikibase-entityid\"}":{"text\/html":{"en":{"P275":" 510:"}},"text\/plain":{"en":{"":"copyright status"}}},"{\"value\":{\"entity-type\":\"item\",\"numeric-id\":50423863,\"id\":\"Q50423863\"},\"type\":\"wikibase-entityid\"}":{"text\/html":{"en":{"P6216":" 619:"}},"text\/plain":{"en":{"":"source of file"}}},"{\"value\":{\"entity-type\":\"item\",\"numeric-id\":66458942,\"id\":\"Q66458942\"},\"type\":\"wikibase-entityid\"}":{"text\/html":{"en":{"P7482":" 415:"}},"text\/plain":{"en":{"":"Wikimedia username"}}},"{\"value\":{\"entity-type\":\"property\",\"numeric-id\":2699,\"id\":\"P2699\"},\"type\":\"wikibase-entityid\"}":{"text\/html":{"en":{"":" 403:"}},"text\/plain":{"en":{"":"creator"}}},"{\"value\":{\"entity-type\":\"property\",\"numeric-id\":2093,\"id\":\"P2093\"},\"type\":\"wikibase-entityid\"}":{"text\/html":{"en":{"":" 615: 419:"}},"text\/plain":{"en":{"":"URL"}}},"{\"value\":\"https:\\\/\\\/commons.wikimedia.org\\\/wiki\\\/User:Cmglee\",\"type\":\"string\"}":{"text\/html":{"en":{"P2699":" 407:"}},"text\/plain":{"en":{"":"author name string"}}},"{\"value\":\"Cmglee\",\"type\":\"string\"}":{"text\/html":{"en":{"P2093":"Cmglee","P4174":" 795:(circles) to the opposite vertex split the triangle into 6 regions. 2. Each region is rotated 90 degrees outwards about its hinge, yielding 3 congruent triangles with sides equal and perpendicular to the line segments between the original triangle's vertices and its centroid (2 155:(circles) to the opposite vertex split the triangle into 6 regions. 2. Each region is rotated 90 degrees outwards about its hinge, yielding 3 congruent triangles with sides equal and perpendicular to the line segments between the original triangle's vertices and its centroid (2 272:– You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use. 408: 620: 559: 555: 511: 420: 616: 507: 416: 412: 404: 551: 400: 328: 562:"}},"text\/plain":{"en":{"P275":"GNU Free Documentation License, version 1.2 or later"}}}}": --> 423:"}},"text\/plain":{"en":{"P2699":"https:\/\/commons.wikimedia.org\/wiki\/User:Cmglee"}}}}": --> 69: 65: 61: 57: 53: 47: 326:; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled 106: 78: 646: 602: 585: 537: 763:
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Visual proof of Lee Sallows's triangle theorem by CMG Lee. 1. Lines from midpoints of edges of length
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Visual proof of Lee Sallows's triangle theorem by CMG Lee. 1. Lines from midpoints of edges of length
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807:). 3. The process can be repeated on each smaller triangle. 4. Each resulting triangle has sides 167:). 3. The process can be repeated on each smaller triangle. 4. Each resulting triangle has sides 114: 700: 695: 42: 316:
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File:Sallows triangle theorem.svg
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GNU Free Documentation License
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