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Triangle ABC is isosceles.

Find angle x.

post-1048-1221180554_thumbgif

Do not assume the figure is drawn to scale.

Do not use the law of sines.

Use simple geometric reasoning, like


  • Isosceles triangles have two equal angles and sides.
  • The angles of a triangle sum to 180 degrees.
  • Things like that.

Have fun.

You can add a line or two if you think it will help.

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There's a hint in the OP hidden in the spoiler tag. I want to add some discussion to that, but I'll hide it as well. I'm guessing the problem requires using the hint to solve.

There are a few lines I've considered adding. One is to add another point F on leg AC so that AFEB is a trapezoid. This hasn't helped me much. I've also tried duplicating the interior lines (BD, DE, and AE) in mirror image, but that hasn't helped much either. I think you might need to add some elevations, maybe in triangle DCE. I'm working on other options (but also real work that I get paid to do, so expect delays).

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There's a hint in the OP hidden in the spoiler tag. I want to add some discussion to that, but I'll hide it as well. I'm guessing the problem requires using the hint to solve.

There are a few lines I've considered adding. One is to add another point F on leg AC so that AFEB is a trapezoid. This hasn't helped me much. I've also tried duplicating the interior lines (BD, DE, and AE) in mirror image, but that hasn't helped much either. I think you might need to add some elevations, maybe in triangle DCE. I'm working on other options (but also real work that I get paid to do, so expect delays).

but so far, none of the lines I've added tell me anything I don't already know without the lines

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x=90?

oops took too long to do a look over...yeah i'm pretty sure x =90°

Well, I don't know the right answer yet, but I can prove this one is incorrect.

Draw a line parallel to AB that passes through E. The point where the new line crosses AC is labeled as F. We know that point F is closer than D is to point C along line AC because angl EAB is greater than angle DBA.

Because new line EF is parallel to line AB, we know that angle FEA = EAB = 70. And x is a portion of angle FEA, so x is less than 70.

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Well, I don't know the right answer yet, but I can prove this one is incorrect.

Draw a line parallel to AB that passes through E. The point where the new line crosses AC is labeled as F. We know that point F is closer than D is to point C along line AC because angl EAB is greater than angle DBA.

Because new line EF is parallel to line AB, we know that angle FEA = EAB = 70. And x is a portion of angle FEA, so x is less than 70.

interesting because the angles i came up with were this

Angles:

ABC = 80°

ABD = 60°

ACB = 20°

ADE = 80°

ADB = 40°

AEB = 30°

AEC = 150°

AED = 90° also angle x

A_B = 50°

A_D = 130°

BAC = 80°

BAD = 10°

BAE = 70°

BDC = 140°

BDE= 40°

BED = 120°

B_E = 130°

CDE = 100°

CED = 60°

DBE = 20°

D_E = 50°

if you could...point out which are incorrect please :D

Edited by Impervious
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Cherry Lane ignore my old angles :P

you're right. I just plugged in the numbers.

And I was just in the process of responding to that, too.

I haven't gotten far enough to comment on the current proposed solution. It passes the reasonableness test, but I'm missing the way to the solution.

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And I was just in the process of responding to that, too.

I haven't gotten far enough to comment on the current proposed solution. It passes the reasonableness test, but I'm missing the way to the solution.

I actually move away from triangles to calculating trapezoid ADEB

while using the information I knew from

ADB

ABC

BAC

BED

while also knowing what the angle D_E is so it gave me the number that was needed in order to make ∆D_E

Then i began to plug in the obvious number that could solve the puzzle (well incorrectly the first time, i got lost in all my markings)

So yeah, there ya go.

(angles were left out for the sake of those wanting to solve it themselves :))

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I actually move away from triangles to calculating trapezoid ADEB

while using the information I knew from

ADB

ABC

BAC

BED

while also knowing what the angle D_E is so it gave me the number that was needed in order to make ∆D_E

Then i began to plug in the obvious number that could solve the puzzle (well incorrectly the first time, i got lost in all my markings)

So yeah, there ya go.

(angles were left out for the sake of those wanting to solve it themselves :))

ADEB is not a trapezoid.

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I actually move away from triangles to calculating trapezoid ADEB

while using the information I knew from

ADB

ABC

BAC

BED

while also knowing what the angle D_E is so it gave me the number that was needed in order to make ∆D_E

Then i began to plug in the obvious number that could solve the puzzle (well incorrectly the first time, i got lost in all my markings)

So yeah, there ya go.

(angles were left out for the sake of those wanting to solve it themselves :))

What is angle BED and how did you get you come by the number?

Edited by Prof. Templeton
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