## Friday, April 4, 2014

### Geometry Problem 1000. Scalene Triangle, Isosceles, Angle, 120 Degree, Midpoint, Distance, Equilateral, Congruence, Perpendicular, Metric Relations

Geometry Problem.

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 1000. 1. CD=AE=2GH=8, GH=4, JH=2.
By problem 998, FH=4.
Hence, x=√12=2√3.

2. http://s22.postimg.org/iff0mclox/problem_1000.png

Let P is the midpoint of DE
As the result of problem 998 we have
Triangle DBC congruent to ABE=> AF=DC=8
And triangle PFH and FHG are equilateral
Since G and H are midpoints of AC and EC => GH= 1/2AE=4
In equilateral triangle FGH altitude FJ= GH.sqrt(3)/2= 2.sqrt(3)

3. Following on from my proof of Problem 998, FGH is equilateral too and is of side 4 so
x = sqrt(16 - 4) = 2sqrt3

Sumith Peiris
Moratuwa
Sri Lanka

4. Tr. ABE and CBD are congruent (SAS) => AE=8
Since G and H are mid-points of AC and EC => GH//AE and GH=4 -----(1)
Similarly GF//CD and GF=4 -----------------(2)
As ABD is isosceles=> m(BDF)=30 and Tr.DBF is 30-60-90 and BF=DB/2.
Similarly BH=EB/2
Tr. BFH similar to ABE with each side of BFH half of corresponding sides of ABE
Hence HF=AE/2=4 ---------(3)
From (1) ,(2) & (3) GFH is an equilateral triangle with each side = 4 units
=> FH=2Sqrt(3)