MHB Congruence of ACB & CED: Similarity Chapter

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ACB and CED are confirmed to be congruent, while the term "similarity" is deemed unnecessary in this context. Triangle ABC is established as a right triangle, and although triangle CED's right angle status isn't explicitly stated, it is implied for the problem's validity. The congruence of angles ACB and CED leads to the conclusion that angles CAB and DCE are also congruent, resulting in angle ACE being a right angle. Consequently, triangle ACE is identified as a right triangle with legs measuring 7 and 24, allowing for the calculation of length AE using the Pythagorean theorem. This discussion emphasizes the importance of congruence in understanding the relationships between the triangles involved.
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Only use that ACB and CED are congruent, not any other symbols in the diagram.
 
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I don't think "similarity" is useful here (and the two triangles are definitely NOT "congruent". We are given that triangle ABC is a right triangle. (It is not specifically said that triangle CED is a right triangle but if angle CDE is not a right angle this problem is un-doable.) Angle ACB is congruent to angle CED. Since both triangles are right angles it follows that angles CAB and DCE are congruent and then that angle ACE is a right angle! So triangle ACE is a right triangle with legs of length 7 and 24. The length of AE follows from the Pythagorean theorem.
 

Thread 'Trigonometry problem of interest'

Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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