Mathematica competitions in Croatia

Question

Let ABC be a right triangle with the right angle at vertex C. Let D be a point on the segment AC and let E be a point on the segment BD such that ABC = DAE = AED. Prove that |BE| = 2|CD|. (Lithuania 2010) FIGURE_0

Accepted Answer

Let ABC = DAE = AED = . FIGURE_0 Then aligned BAC &= 90^ - , BAE &= (90^ - ) - = 90^ - 2, BEA &= 180^ - . aligned Applying the law of sines to the triangle ABE: |BE| = BAE AEB |AB| = (90^ - 2)(180^ - ) |AB| = (2) |AB|. Since |CD| = |BC| ctg CDB = |BC| ctg(2), we finally obtain |BE||CD| = (2) |AB||BC| ctg(2) = (2)|AB| |BC| (2)(2) = (2)|AB| |AB| = 2.