Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Sunday, March 22, 2015
Problem 1102: Right Triangle, Incircle, Tangency Point, 90 Degree, Cathetus, Metric Relations
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Let BC=a, AC=b, AB=c, and s=(a+b+c)/2
ReplyDeleteAD=s-c, DC=s-a
DC/BC = DG/AB
(s-a)/a = DG/c
DG = (s-a)c/a
AD×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 AB²
Hence, AB² = 2 AD×DG
slight error on lines #2 to #5: with defs in line #1, AD=s-a and DC=s-c
ReplyDeleteremaining lines are ok
(s - a)(s - c) / ac = cos² (B/2) = 1/2
ReplyDeleteDG = (s - c) tan C = (s - c).c/a = c.(s - c)/a
2.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)/a
ReplyDeleteSo 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 result
In the simplification we have used the fact that b^2 = a^2 + c^2
Sumith Peiris
Moratuwa
Sri Lanka