Friday Puzzler -- "Easy" geometry problem

Hint

Alex Y

The solution requires defining a new point F on one of the existing lines, and then the lines between F and the existing points define more triangles that allow you to determine x.

Isosceles and equilateral triangles are key.

Re: Hint

Larry Barrett

I think F is located on line AC so that line EF is parallel to line AB. Then draw a line from F back to B. It also helps to identify point G as the intersection of lines AE and BD and to determine the four angles around point G.

Re: Hint - not so sure now

Larry Barrett

My answer to the original question - what is angle AED? - is 20 degrees.

The other day I thought I saw the way to the answer by defining point F as stated. Now not so sure; at least, I can't remember how I got from there to my answer.

That's a start

Alex Y

but not the right answer:-(

And an apology--I was unplugged in Glacier National Park last week, so wasn't responsive to your posts.

Creating the line parallel to the base, as you did, is the first step in one of the two solutions I have seen for this puzzle. And the line from F to B is helpful.

Did you mean to say that G was the intersection of AE and BF? That is more interesting.

Re: That's a start

Larry Barrett

'Unplugged in Glacier' could be a movie title (Sleepless in ...). Hope you enjoyed your trip.

I am using H as the intersection of AE and BF. Here is what I have so far:

From the original problem you can determine that:

Triangle ABC is isosceles and the lower angles are 80* (use * as a degree symbol).

Angle ACB is therefore 20*.

Angle ADB is 50*.

Angle BEA is 40*.

Angle AEC is 140* and angle BDC is 130*.

Now draw a line parallel to AB from E to the opposite side and lable the intersection F. Triangle FCE is also isosceles with lower angles 80*.

Therefore angle FEA is 60*.

From the original problem, triangle AEC is also isosceles since angle CAE and angle ACE are both 20*. Therefore triangle BFC is also isosceles and congruent with triangle AEC.

Trapezoid AFEB is isosceles (if there is such a term for trapezoids) because sides AB and FE are parallel and angles FAB and EBA are both 80* and angles AFE and BEF are both 100*. Diagonal AE and the new diagonal BF form congruent interior triangles AHF and BHE (H is the point where the diagonals intersect.)

So triangle FHE is isosceles with sides FH and EH equal. And we know that angle FEA is 60*; therefore angle EFB is also 60* and triangle FHE is equilateral.

An Alternate Tack

Alex Y

Larry has Identified a point F, such that EF is parallel to AB. That is the key to one solution. Everything else is labeling points and creating lines between points, and working with properties of triangles. (It's harder than that makes it sound ;-) )

Here is another point that could be used for a different solution. I'll use "M" to make sure not to confuse with the solution Larry is following.

M is a point on BC such that MAB is 20*. From there, it is all about identifying isosceles and equilateral triangles, and line segments of the same length.

Next line

Alex Y

You need a line from D to H.

Re: Next line

Larry Barrett

I think I need a protractor. You weren't kidding when you said this 'easy' problem was really hard. I looked at the other approach, too. Easy to find more isosceles triangles and an equilateral, and equal line segments, but still haven't discovered DEA.

Yes, it is :-)

Alex Y

I'll post the alternative solution, and it might give you hints to this one.

Solution

Alex Y

Here are the steps to the answer with the alternative starting point. In the diagram below, all the colored line segments are the same length, as explained below the diagram.

Legs of a 80-20-80 isosceles triangle

Legs of a 50-80-50 isosceles triangle

Legs of a 40-100-40 isosceles triangle

Sides of an equilateral triangle

So the triangle with one yellow and one blue leg is isosceles, and the angle near the bottom of the diagram is 180-80-60=40, so the triangle is 70-40-70

The angle we want is 70 from the blue-yellow triangle minus 40 from the blue triangle = 30*

Easy, huh? Not so I can tell it!

Re: Next line - you didn't say ....

Larry Barrett

That you need colored pencils to solve this one. Every sketch I make ends up with too many lines, angles, numbers, etc, plus not to scale.

After drawing line DH I can see that ADH is isosceles 20 80 80. So lines AD, AH, AB, and BH are all equal.

Also, angle DHF is 40, angle HFD is 40 so HFD is isosceles 40 40 100 and lines FD and HD are equal.

But I still quite see how to get angle DEA.

Compare

Alex Y

triangles DFE and DHE.

Re: Compare

Larry Barrett

Well, with enough hints I think I finally figured it out.

We know that triangle FHE is equilateral so FE=HE=FH and angle FEH is 60*.

We also know that DF and DH are equal because triangle DFH is isosceles .

So triangles DHE and DFE are congruent, which means angle FED and DEH are equal. The sum of FED and DEH is equal to FEH = 60*. So angle DEH is 30*.

I see, said the blind man as he picked up his hammer and saw.

Re: Compare

Alex Y

LOL! It amazing how hard this is, even after getting the added point. Beyond getting that point, there is nothing there that we didn't learn in 9th grade geometry, but it is still VERY hard!

Nice job thinking of the line parallel to the base as the start!

Now work on the handshakes. No math there -- just pure logic.