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.

ReplyDelete