## Friday, July 13, 2018

### Geometry Problem 1363: Square, Angle Bisector, Perpendicular, 45 Degrees, Angle, Sketch, iPad Apps

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Details: Click on the figure below..

1. BG = BC (BF: bisector of angle CBE; FG perpendicular to BE)
So BG=BC= AB; (square ABCD)
So A,G,C on same circle with center at B;
Connect AC,GC.
Angle CAG = ½ angle GBC =angle GBF=angle FBC.
Because BG=BC= AB;
So angle BAG =angle AGB;
Angle BAG =45+ angle GBF ;
Angle AGB =angle GBF + angle AHB;
So angle AHB=45.

2. let m(ABE)=x
=> m(GBH)=45-x/2 -------(1)
Since m(FGB)=m(FCB)=90 and BF bisector of m(CBE)
=> BCFG is concyclic and BG=BC
Triangle ABG is isosceles and m(BAG)=90-x/2----------(2)
From (1) and (2), m(H)=180-(90-x/2)-(x+45-x/2)
=>m(H)=45

3. triangle BGF = triangle BCF
=> BC=BG=BA => BAG isosceles

let alpha=angle(ABE)
Consider triangle ABH :
angle(H) = 180 - angle(A) - angle(B) = 180 - (alpha+(90-alpha)/2) - (180-alpha)/2) = 45°

4. Florin Popa, Comănești, România.
∆BGF≡∆BCF,=>GF ≡CF,BG ≡BC(1)
Ducem cercul cu centrul in B și rază BC, care trece prin punctele A și G. BF taie cercul in punctul P iar prelungirea lui BF taie cercul in M. Ducem diametrul CBN.
Din (1) rezultă că arcele (CP) ̂=(GP) ̂. Se arată ușor că ∆BMN≡∆BGP de unde rezultă că (MN) ̂=(GP) ̂.
Atunci unghiul AHB = (Arc AM-arc GP)/2 = (arc AM – arc MN)/2 = arc AN /2 = 90/2=45.

5. Is CF=FG,BC=BG=AB => β-α=45.But <BHG=<BGA-<GBH=β-α=45.