Proposed Problem

Click the figure below to see the complete problem 406 about Right triangle, 15 degrees, Midpoints, Congruence.

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Complete Problem 406

Level: High School, SAT Prep, College geometry

## Wednesday, December 23, 2009

### Problem 406. Right triangle, 15 degrees, Midpoints, Congruence

Labels:
15 degree,
congruence,
midpoint,
right triangle

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Join G to B

ReplyDelete|GB|=1

Join F to B

|FB|=1

see m(FBG)=60 and FBG is equilateral triangle

|FG|=1

draw BM perpendicular to FG, GN perp to AC

ReplyDeleteRight tr BMG and GNC are congruent ( BG=GC)=> GN = x/2

but GN = 1/2 (ang 30) => x = 1

Connect BF and BG, AD=CE=2; AD and CE both hypotenuses. BF and BG are medians for them, so BF=BG=1. The median reveals Angle ABF=Angle CBG=15 Degrees. Angles (ABF+ABG+CBG)=90 Degrees, so Angle ABG=60 Degrees, which means FBG is equilateral.

ReplyDeleteTherefore, FG=x=1.

Solved in a very complex manner but Newzad's solution is so simple.

ReplyDeleteSee the

ReplyDeletedrawing∠BAD= π/2 => B, A and D are concyclic, circle with center F middle of BD

=> FB=FA=FD and ∠BAF=15°

In the same way, GC=GA=GE and ∠CAG=15°

=>∠FAG=90-15-15=60°

ΔFAG is isosceles in A (AF=AG) and ∠FAG=60° => ΔFAG is equilateral

Therefore x=FG=BD/2=CE/2AD=CE (given)

ReplyDeletetriangle BAD congruent to triangle triangle BCE (AAS)

=>BA=BC & BD=BE

AE=BA-BE

CD=BC-BD

=>AE=CD

Let intersecting point of AD & EC be H

AE=CE (proved)

triangle EAH congruent to triangle DCH (AAS)

=>AH=CH

<HAC=<HCA

<EAC=<DCA=45

In triangle ADC

<ADC=105

sin105/AC=sin45/2

AC=2sin105/sin45

AC=2sqrt2*sin105

AC=sqrt3+1

In triangle AHC

<AHC=120

sin120/AC=sin30/AH

AH=ACsin30/sin120

AH=AC/sqrt3=(sqrt3+1)/sqrt3

Triangle FHG ~ Triangle AHC

x/AC=FH/AH

x/AC=(AH-1)/AH

x/AC=1/(sqrt3+1)

x=1