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

Labels:
circle,
cyclic quadrilateral,
inscribed,
secant,
similarity,
triangle

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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)