Problem 12
Let ABC be any triangle. The internal angle trisectors of A, B and C, meet at D, E, and F respectively, as shown in the simulation below. Then, prove that:
ΔDEF is equilateral
Here’s a simulation that demonstrates the problem.
You can drag the vertices A, B, and C and observe ΔDEF. Is it always equilateral?
Solution
I won’t provide the full solution here, but only give you some cues that’ll help you solve the problem yourself.
While the problem could be solved using pure geometry, the hints I provide here would involve trigonometry.
Here’s what you’ll need to know to be able to solve the problem.
Sine Rule
In any triangle ABC,
a/sinA = b/sinB = c/sinC = 2R
where,
a, b, and c are the lengths of BC, CA, and AB
R is the circumradius
Cosine Rule
In any triangle ABC,
cosA = (b² + c² – a²)/2bc
cosB = (c² + a² – b²)/2ca
cosC = (a² + b² – c²)/2ab
where,
a, b, and c are the lengths of BC, CA, and AB
Now, here’s how you can proceed:
First, you can find BD by applying the sine rule in ΔABC and ΔABD.
Next, you can find BE by applying the sine rule in ΔABC and ΔBEC.
Finally, you can apply the cosine rule in ΔBED to find DE.
Well, that’s all that you need to know here. 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.
Problem Source: CutTheKnot
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