## Friday, October 10, 2008

### Elearn Geometry Problem 188

See complete Problem 188 at:
www.gogeometry.com/problem/p188_square_45_degrees_problem.htm

Square, Diagonal, 45 Degrees Angle. Level: High School, SAT Prep, College geometry

1. O is the midpoint of GH, EO cuts CD at E', FO cuts BC at F'
⊿AGH'≡⊿AGH (SAS), ∠BGA＝∠AGH and ∠GHA＝∠AHD
E,A,D,H are concyclic, HE⊥GA, so is GF⊥AH
G,E,F,H are concyclic with center O
OG＝OE, ∠OEG＝∠EGO＝∠BGE, EE'⊥CD, so is FF'⊥BC
a²+b²＝2(CE'²+CF'²)＝2(x/2)² ⇒ x²=2(a²+b²)

2. Here's another way to do this: Let A,B,C & D be (0,0),(0,s),(s,s) and (s,0). Also let BG =t. We can, therefore, say that: AG is y=sx/t and BD:x+y=s. Hence, E is (st/(s+t),s^2/(s+t)) Thus, BE^2= a^2 = (st/(s+t))^2 + (s-s^2/(s+t))^2=2[st/(s+t)]^2. Now let DH =q. Hence slope AH=q/s but the angle between AG and AH is 45 deg. Hence [s/t -q/s]/[1+sq/ts]= tan(45)=1. Hence slope DH: (s-t)/(s+t). We locate be F and determine DF^2=b^2 - (s-t)^2. Thus 2(a^2+b^2)= (2st/(s+t))^2 +(s-t)^2 = (s^2+t^2)^2/(s+t)^2 ----(1)
While x^2 =(t-s)^2+(s-s(s-t)/(s+t))^2=(s^2+t^2)^2/(s+t)^2 ----(2).
By (1) & (2) we can say that x^2=2(a^2+b^2)
Ajit: ajitathle@gmail.com

3. can i do a question ,please

4. CH=HD is not an hypothesis.

5. Charxith said...

Let point H be on point D and point G on point C, having GH = DC, AG = BD, and angle CAD = angle GAD. This means that x = DC and that b^2 = 0.
This means x^2 = 2a^2, and since a = AC/2, 2a^2 = a^2 + a^2, segments AC and BD intersect at center of square creating 4 isosceles right tri,

we can conclude that a^2 + a^2 = x^2, or 2a^2 = x^2, by pythagorean theorem.

6. Charxith said...

Therefore from my comment before, x^2 = 2(a^2 + b^2), since b^2 = 0.

7. Join AC and let m(HAD)=x
=>m(CAH)=45-x, m(GAC)=x and m(BAG)=45-x
Observe that m(BAE)=m(CAH)=45-x, m(ABE)=m(ACH)=45
=> ABE and ACH are similar triangles
=> AB/BE=AC/CH
=> CH=Sqrt(2)a -----------(1)

Similarly the triangle ADF and ACG are similar and GC=Sqrt(2)b--------(2)
Applying Pythagorean to triangle GCH

=> GH^2=CH^2+GC^2
=> GH^2=2(a^2+b^2)
Q.E.D