Mind your decisions, I'm Presh Talwalkar
In triangle ABC
Construct BD so that a BD is equal to 60 degrees and DBC is equal to 20 degrees
Then construct AE so that BAE is equal to 70 degrees an EAD is equal to 10 degrees
finally construct DE
the question is to solve for angle AED
marked as X
You're supposed to solve this using only elementary methods. That means no trigonometry.
I admit, I could not solve this problem and I had to read the solution online which I will present in the rest of the video
Pause if you'd like to give this problem a try and when you're ready keep watching to learn how to solve this problem
We'll get started by focusing on triangle ABE
Angle B is equal to 80 degrees which means the remaining angle in this triangle is equal to 30 degrees
We'll then go back to the original diagram
Construct DF so that DF is parallel to AB
We'll proceed in the following steps. We'll first show that angle DFE is equal to 80 degrees
That'll be the easy part. We then need to show that FD is equal to FE. That's the hard part
From there, we'll be able to solve that X is equal to 20 degrees. So let's get started
We'll first consider just the following similar triangles
The reason they're similar is because DF was constructed to be parallel to a B
So angle C is equal to 20 degrees which means the remaining angles will be equal to 80 degrees
We've completed the first part of the proof
Well now go to the hard part of the proof
Well first consider the supplementary angles, which will be equal to 100 degrees each
now again, let's construct BD which divides this angle into 60 degrees and 20 degrees in
This triangle the remaining angle must be equal to 60 degrees
From here. Let's construct a F which intersects at point G
Notice that this triangle will exactly be congruent to this triangle
Therefore angle a FD will be equal to 60 degrees
the remaining angle and triangle FGD must also be equal to 60 degrees and then by vertical angles
We have a 60 degree angle here
Then in triangle a B G the remaining angle must be equal to 60 degrees as well
We constructed a couple of equilateral triangles
Furthermore angle FAD must be equal to 20 degrees because angle a as a whole is equal to 80 degrees
We've thus constructed an isosceles triangle as follows
The sides opposite the 20 degree angles must be equal to each other.
So AF must be equal to FC
Now return to the diagram and then let's draw in AE again
Recalling that we had a 10 degree angle here from the original information
The remaining angle must be equal to ten degrees so that we have a total sum of 20 degrees
We're then going to bisect angle C which will intersect at Point G
These will be two ten degree angles. We now have a couple of congruent triangles
Notice the triangle AGC is exactly congruent to triangle CEA
Therefore AG is exactly equal to CE
What this means is the remaining segments FG and FE will have the same length
From here we go back to the equilateral triangle of FGD. And since it's equilateral we must have FG
equal to FD
Consequently FD is equal to FE and we prove in this part
Now, how do we solve the problem?
We go back to the original diagram and recall that we calculated that angle DFE is equal to 80 degrees
so we'll now focus on triangle DFE
We have an isosceles triangle or one angle. The vertex angle is equal to 80 degrees
The two equal angles must be equal to 30 degrees plus X
So we have the following equation 2 x 30 plus X plus ad is equal to 180
We subtract 80 from both sides
we then divide by 2 and
then we simplify and subtract both sides by 30 and we get that X is equal to 20 degrees and
That's the answer. It was a long journey, but it was a satisfying one
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