Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 805.
Saturday, September 22, 2012
Problem 805: Stewart's Theorem, Triangle, Sides, Cevian, Metric Relations, Measurement
Subscribe to:
Post Comments (Atom)
In triangle ADC,
ReplyDeleteb^2 = x^2 + m^2 - 2 x m cos(ADC)
In triangle BDC,
a^2 = x^2 + n^2 - 2 x n cos(BDC)
Since cos(ADC) + cos(BDC) = 0,
(x^2 + m^2 - b^2) / (2xm) + (x^2 + n^2 - a^2) / (2xn) = 0
n(x^2 + m^2 - b^2) + m(x^2 + n^2 - a^2) = 0
a^2 m + b^2 n
= (m+n) x^2 + mn(m+n)
= x^2 c + cmn