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− | Given , it follows that . Extend to intersect at , making . Triangles and are similar. Also, and , which implies points , , , and are concyclic.
| + | The Problem shows that , it follows that . Extend to intersect at , we get . Making triangles and similar. Also, and , which points , , , and are concyclic. |
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− | Furthermore, . Triangle is congruent to , and . Let , then points , , , , and are concyclic.
| + | And . Triangle is congruent to , and . Let , then points , , , , and are concyclic. |
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− | It is also given that and . Additionally, and with , making a rhombus. Consequently, and triangle is congruent to , while is congruent to which is congruent to , and .
| + | Finally and . and with , making a rhombus. And and triangle is congruent to , while is congruent to which is congruent to , so . |
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| ~Athmyx | | ~Athmyx |
Revision as of 11:32, 20 April 2024
Problem
Triangle has a right angle at . Let be the point on line such that and lies between and . Point is chosen so that and is the bisector of . Point is chosen so that and is the bisector of . Let be the midpoint of . Let be the point such that is a parallelogram. Prove that and are concurrent.
Solution
The Problem shows that , it follows that . Extend to intersect at , we get . Making triangles and similar. Also, and , which points , , , and are concyclic.
And . Triangle is congruent to , and . Let , then points , , , , and are concyclic.
Finally and . and with , making a rhombus. And and triangle is congruent to , while is congruent to which is congruent to , so .
~Athmyx
See Also