Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Saturday, December 12, 2015
Geometry Problem 1172: Cyclic Quadrilateral, Circle, Secant, Triangle, Similarity, Metric Relations
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Let EC = y.
ReplyDeleteFrom similar triangles, we have
x/(y+c) = y/(x+a) = b/d
dx = by + bc
dy = bx + ab
dx - by = bc
bx - dy = -ab
By solving, we have
x = b(ab + cd)/(d² - b²)
One of the easiest for some time Antonio!
ReplyDeleteAlso 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)