Problem 6
ABCD and BEFG are congruent squares, such that 0° ≤ ∠CBE ≤ 90°. Then, prove that:
AE = GC + DF
Here’s a simulation that demonstrates the problem.
You can drag the the point E to rotate the square BEFG. Is AE always equal to DF + GC?
Solution
I won’t provide the full solution here, but only give you some tools and hints that’ll help you solve the problem yourself.
Here’s you’ll need to know to be able to solve the problem.
Similar Triangles
If the corresponding angles of two triangles are equal, then they are said to be ‘similar’. That is, they have the same shape. This also means that their corresponding sides are proportional.
For example, if ∆ABC and ∆DEF are similar, then
A = D, B = E, C = F
AB/DE = BC/EF = CA/FD
Here’s a simulation that shows two similar triangles. You can drag the vertices and observe the angles and side lengths.
To prove that two triangles are similar, you can prove that their corresponding angles as equal. Then, you can use the fact that their corresponding sides are proportional.
Now, to proceed, you have to find some similar triangles in the figure, such that their sides are AE, CG, and DF (because, only then you’ll be able prove a result that involves them).
That’s all for now. I suggest you proceed on your own from this point. Good luck!
Were you able to solve it using the above hints? Or, did you solve using any other method? Or, do you need some more help? You can reach out at the Facebook page.
Further Research
What happens if ∠CBE > 90°? Will we get the same result? If yes, can you prove it?
If not, will some other relation always hold between AE, GC, and EF? Again, how can you prove that?
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