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Question

Let ABC be a triangle where M is the midpoint of line AC, and line CN is the angle bisector of angle ACB with N on line AB. Let X be the intersection of the median BM and the bisector CN. In addition, triangle BXN is equilateral with AC=2. What is BX^2?

Answer 1

This problem has used 60^o-30^o-90^o special triangle rules. The starting point is ΔBXN is an equilateral triangle.

Given: ΔBXN is an equilateral triangle => BX=BN=XN and all the angles measure 60^o

CN= 2 XN , as X is the bisector of CN

=>CN=2BN as XN=BN

In ΔNCB, CN =2BN, Included angle∠N = 60^o

Comparing with 60-30-90 special triangle, by SAS similarity ΔBCN is a 60^o-30^o-90^o triangle, with ∠B = 90^o

and ∠BCN=30^o -------(1*)

=>∠BCA = 2∠BCN [since CN is the angle bisector]

=> ∠BCA = 2 ×30^o= 60^o

In ΔABC, ∠B = 90^o, ∠C= 60^o, AC is hypotenuse

As AC = 2. BC = AC/2 = 1 [ Using rule for 60^o-30^o-90^o special triangle]

Applying the rule for 60^o-30^o-90^o special triangle on ΔNCB, BC/CN=√3 /2 , CN = BC×(2/√3) = 2/√3

BX=XN =CN/2 = 1/ √3 [ X being the mid point of CN)

=>BX²= 1/3

I hope this helps. I apologize for being able to upload a drawing.

(1*)You can use cosine law of triangles to derive (1) as well.

Answer 2

This was a long involved process, but I found a formula someone had posted online that with an equilateral triangle your formula to find the distance from the centroid to the vertex is 2/3* (√3*x)/2 where x is the length of one of your triangle sides. And when I plug in 2 for x into that formula, I get the same answer as I did in the video so maybe that formula was taught to you and that's all you had to do in which case the process would've been a lot quicker. But that's a semi-longer way with the reasoning behind it (I had never seen that formula before so I'm not sure if you have). Hope this helped!