## Sunday, March 22, 2015

### Problem 1102: Right Triangle, Incircle, Tangency Point, 90 Degree, Cathetus, Metric Relations

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. 1. Let BC=a, AC=b, AB=c, and s=(a+b+c)/2

DC/BC = DG/AB
(s-a)/a = DG/c
DG = (s-a)c/a

= (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

2. slight error on lines #2 to #5: with defs in line #1, AD=s-a and DC=s-c
remaining lines are ok

3. (s - a)(s - c) / ac = cos² (B/2) = 1/2
DG = (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

4. AD = c - r where r is the inradius. From similar Tr.s DG = c(a-r)/a

So 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