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

Post your solutions or ideas in the comments.

## Friday, October 10, 2008

### Elearn Geometry Problem 188

Labels:
45 degrees,
angle,
diagonal,
square

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O is the midpoint of GH, EO cuts CD at E', FO cuts BC at F'

ReplyDeleterotate ⊿ADH about A such that AD and AB coincide and H maps to H'

⊿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)

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)

ReplyDeleteWhile 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

can i do a question ,please

ReplyDeleteis CH=HD ?

ReplyDeleteCH=HD is not an hypothesis.

ReplyDeleteCharxith said...

ReplyDeleteLet 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.

Charxith said...

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

Join AC and let m(HAD)=x

ReplyDelete=>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