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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1172.
Let EC = y. From similar triangles, we havex/(y+c) = y/(x+a) = b/ddx = by + bcdy = bx + abdx - by = bcbx - dy = -abBy solving, we havex = b(ab + cd)/(d² - b²)
One of the easiest for some time Antonio! Also EC = b(ad+bc)/(d^2-b^2)We can write similar expressions for the diagonals again using similar triangles and deduce using Ptolemy that BD^2 = (ac+bd)(cd+ab)/(ad+bc)AC^2 = (ad+bc)(ac+bd)/(ad+bc)