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Parallel Midpoints (Posted on 2016-07-30) Difficulty: 4 of 5

Let BD be an altitude of ΔABC with D between A and C (not the midpoint).
The bisector of ∠ACB intersects BD and AB in points P
and Q respectively. The bisector of ∠CAB intersects BD
and BC in points R and S respectively. F and G are the
midpoints of PQ and RS respectively.

Prove that FG is parallel with AC if and only if ∠ABC = 90°.

See The Solution Submitted by Bractals    
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not full solution Comment 1 of 1
We will measure the altitude of F and G points from AC, to see when they are the same. 

To easy things I use as notation h=BD and hf, hg, hs, etc, the altitude of points F, G, S, etc. 
Also the angle BAC is 2a and ACB is 2b.


For hp
tan.2b=h/DC   tan.b=hp/DC    From here hp=h*tan.b/tan.2b and so:


For hq

CQ/sin.2a=AC/sin.(2a+b)  AC=h*sin.(2a+2b)/(sin.2a*sin.2b)  so that CQ=h*sin.(2a+2b)*sin.2a/(sin.2a*sin.2b*sin.(2b+a))


To avoid excessive complesity in the text here we are already going to suppose that the angle ABC is 90°. If the enunciate of the puzzle is right we should get hf=hg

With ABC angle = 90° then 2a+2b=90 and 2a+b=90-b. 

The expression for hq is then
hq=h*sin.b/sin.2b*cos.b   hq=h/2(cos.b)^2

Then hf=1/2(hp+hq)= h*(1-(tan.b)^2)/4 + 1/4(cos.2b)^2)
Multiplying and dividing for (cos.b)^2 the first summand    

and directly from here 


But for hg all the formulas are simmetric, changing a with b and b with a. and using the correspective sides of the triangle. 
So then: 


Edited on August 3, 2016, 1:10 pm
  Posted by armando on 2016-08-03 09:58:40

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