All that part of Step 4 from 05:44 to 08:56 is unnecessary, because you are basically proving the perpendicular bisector theorem. So you might as well use that theorem directly, by observing that CD is congruent to CF (the legs of the isosceles DCF), so the point C is equidistant from DF's endpoints D and F, and similarly, since we have proven that DGF is equilateral and DG is congruent to GF, then G is equidistant from DF's endpoints D and F as well. This means that CG is a perpendicular bisector of DF, bisecting it into two equal halves at DH and HF at H, where it is perpendicular to it. And from the same reasons, it bisects the angles DCF and DGF.
Much easier way, no extra lines, nothing but + and -. All you need to know is the angles of triangle ABC and of BE and the crossover of the to lines in the middle, then the large triangle from the top corner to the bottem right corner is isosilies, so the line going across the middle is lined up to cut 2 similar triangles, they both have the same angles, therefore transfer them to the upper triangle and you are left with the answer: 20
This problem is different from the problem described by Mindyourdecisions. The 80 degree angles are divided differently. Can your problem be solved using the Mindyourdecisions method?
The web site it was posted on says this problem has a fairly easy solution. I would assume there has to be an easier solution than this. Thank you anyway.
it seems like u assuming it does bisecet the triangle dgf to get a 30 60 90 triangle.. but what if it didnt bisect the triangle u cant say the point are collinear. so we are to assume the triangle dgf are right in the middle and has to be bisected from observation? or any reason other then observation that we an assume that it did in fact get bisected?
im thinking maybe by reason of construction u are forcing segment cg to bisect df and triangle dgf .. i know it work out because of the 90 degree it formed but im not really sure why it has to go right down the middle unless u forcing it to be by way of construction.
Vin Lee it is not forced; point H belonging to DF is chosen s.t. angle DCH= angle FCH. Then, triangles DCH and FCH are congruent because there is the common side CH, CD=CF (shown at a previous step) and the angle between them (10 deg); so the triangles are congruent so DH=HF; then you do same thing with GH' and you will get that DH'=H'F; if H is in the middle of DF and H' is also in the middle of DF it means that they are one and the same point; the angles there are 90 deg, that comes from 'sum of the angles in a triangle is 180 deg'; so the points are coliniar, nothing is forced; i hope it clarified it; AGAIN: You assume it bisects and then you prove that it falls in the middle and forms a 90 deg angle. NOT the other way around!
+Vin Lee There is a theorem that when you have two line segments (let's say A and B) which are congruent (that is, of the same length, A=B), and from each one of them you subtract another pair of congruent segments (let's say, C and D, where C=D), then their differences (that is, what's left from each of them: X=A-C, Y=B-D) are also congruent (X=Y). Or in short: The differences of congruent segments subtracted from congruent segments are congruent. Or in shorter: If A=B, and X=A-C, Y=B-D (where C=D), then X=Y.
I am not sure if it's the problem you are referring to, but I recently posted a video on the World's Hardest Easiest Geometry Problem No. 3 , which has base angles of 10/70 and 50/30. You need a compass and straight edge to solve it.
u should do some annotation instead of just verbally dictating. it will be easy to follow instead of just have to pause and rewind to catch up what me miss cause u going to fast
