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(October 27, 2018 at 11:39 am)Grandizer Wrote: Ok, I'm about to give up on those geometry problems I linked to a few pages ago here. But let me try doing the first problem in stages first and get feedback along the way here.
This stage is very easy, but let me know if I made any calculation errors here:
I see a 60 degree angle in one of the smaller triangles. Maybe if I force an equilateral triangle in there, will that lead me somewhere? Also, the main triangle is isosceles. Hmmmmmm ... ... ...
Yes, make other shapes inside. ... And outside if it helps you.
Ok, unfortunately, I went and looked for the solution (to the first one).
Not exactly neat or orderly, and the explanations I saw on YouTube were horrible in terms of presentation, so what I will do later is present the solution here (under Spoilers) in a "clean" way in stages.
(October 30, 2018 at 6:03 am)Kit Wrote: The Japanese multiply using lines.
That's really cool! I'd heard something about that somewhere along the line, but I'd long forgotten it and I've never actually seen how it works. Thanks for posting that.
Note line DF parallel to AB was drawn. With parallel lines, corresponding angles are equal. Hence, we can know the measures of the angles CDF and CFD. Consequently, we can also know the measures of their respective supplementary angles ADF and BFD. Colors help indicate the exact spaces for some of the depicted angles. Also, some of the measures noted in the earlier image were removed in this one to make sure things are kept relatively neat.
In my opinion, studying mathematics is like learning a second language; it just takes time and patience. Admittedly, some learn faster than others and some are "born" into it, but, everyone can do it.
Note line DF parallel to AB was drawn. With parallel lines, corresponding angles are equal. Hence, we can know the measures of the angles CDF and CFD. Consequently, we can also know the measures of their respective supplementary angles ADF and BFD. Colors help indicate the exact spaces for some of the depicted angles. Also, some of the measures noted in the earlier image were removed in this one to make sure things are kept relatively neat.
Note image clearly not drawn to scale. In this stage, another line was drawn from A to F.
Since triangle CAB is isosceles, and CDF is similar to CAB (therefore, CDF is also isosceles), then the length of DA is equal to the length of FB. You will have to do some algebra to confirm this for yourself. But once you figure that out, you can then deduce that triangle FDA is congruent to triangle DFB (SAS congruency). We then can determine that the measure of angle AFD (being the same as that of BDF) is equal to 60. Therefore, triangle DFG is equilateral, and the measure of angle DGF is 60.
Angle AGB is opposite to angle DGF, therefore its measure is 60. Hence, we can also deduce triangle ABG is also equilateral.
Solution not fully done yet, but I'll get to posting the continuation later.
I had to restore the measures of some of the original angles (angles CAE and EAF) to clarify things here, and hopefully it's clear what angles correspond to which regions as indicated by the matches in colors. If still confusing, let me know. This solution is a bit of a mess, but I've tried to make this as clean as possible.
In this stage, one more line had to be drawn, a line from C to G which just happens to bisect angle ACB (remember that its measure is 20). If not convinced, work out how to prove triangles CAG and CBG are SSS-congruent by remembering that ABG is an equilateral triangle. Should be easy.
Consequently, all new angles in the image above should be easy to figure out on your own, so I won't spend too much energy trying to show how their measures have been determined.
Now the next step is really tricky to visualize, but you can prove that triangles ACE and CGA are ASA-congruent (the angle measures provided in the image are there to help you out).
This means the corresponding sides CE and AG are equal in length (even though its not obvious in the image as it's not drawn to scale).
We also can know CF and AF are equal in length as they are the equal sides of the isosceles triangle ACF (take a look at the base angles if not convinced).
Therefore, CF - CE = AF - AG, which is the same as saying EF = GF.
Now as triangle DFG is equilateral, we know that DF = GF. Therefore, DF = EF. This means triangle DFE is isosceles with base angles DEF and EDF equal to each other.
We can know by looking at the shown angle measures and doing some basic calculation that angle AEF = 30. Therefore, DEF = 30 + x. As it is equal to half the sum of the measures of both base angles DEF and EDF, with the sum being equal to 100 (180 - 80), then the measure of DEF is 50.
Therefore 30 + x = 50 => x = 20
And we finally get our answer.
I love challenges, but this was way too messy for me.
