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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below to enlarge it.
Let BC=a, AC=b, AB=c, and s=(a+b+c)/2AD=s-c, DC=s-aDC/BC = DG/AB(s-a)/a = DG/cDG = (s-a)c/aAD×DG= (s-c)(s-a)c/a= 1/4 (b² - (a-c)²) c/a= 1/4 (2ac) c/a= 1/2 c²= 1/2 AB²Hence, AB² = 2 AD×DG
slight error on lines #2 to #5: with defs in line #1, AD=s-a and DC=s-cremaining lines are ok
(s - a)(s - c) / ac = cos² (B/2) = 1/2DG = (s - c) tan C = (s - c).c/a = c.(s - c)/a2.AD. DG =2.(s - a). c(s - c) / a = 2. [(s - a)(s - c) /ac].c² = 2. (1/2).c² = c² = AB²
AD = c - r where r is the inradius. From similar Tr.s DG = c(a-r)/aSo AD. DG = c(a-r)(a-r)/a Using r = 1/2 (a+c -b) and simplifying we get that AD AG = c^2 /2 which is the required resultIn the simplification we have used the fact that b^2 = a^2 + c^2Sumith PeirisMoratuwaSri Lanka