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# Geometry problem

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1. No idea where to start on this A-Level question from 1972:

The point X divides the side BC of the triangle ABC internally in the ratio m:n, where m + n = 1. Show that AX^2 = mb^2 + nc^2 - mna^2.
2. (Original post by HapaxOromenon3)
No idea where to start on this A-Level question from 1972:

The point X divides the side BC of the triangle ABC internally in the ratio m:n, where m + n = 1. Show that AX^2 = mb^2 + nc^2 - mna^2.
Use cosine rule on each of the smaller triangles, on the angle at X.

Cos AXB = ....

etc.

Notice that AXB = 180 - AXC, hence their cosines are related by....
3. Start with the cosine rule AX^2 = BX^2 + c^2 - 2(BX)(c)cosB

and also cos B = (a^2 + c^2 - b^2) / 2ac

Find a solution for AX^2 and substitute in for n knowing that m+n=1
4. (Original post by ghostwalker)
Use cosine rule on each of the smaller triangles, on the angle at X.

Cos AXB = ....

etc.

Notice that AXB = 180 - AXC, hence their cosines are related by....
They're negatives of each other so we get (m^2a^2+AX^2-c^2)/(2maAX) = -(n^2a^2+AX^2-b^2)/(2naAX)
multiplying by 2aAX gives (m^2a^2+AX^2-c^2)/m = (b^2-AX^2-n^2a^2)/n
multiplying by mn gives n(m^2a^2+AX^2-c^2) = m(b^2-AX^2-n^2a^2)
Thus nm^2a^2 + nAX^2 - nc^2 = mb^2 - mAX^2 - mn^2a^2
so (n+m)AX^2 = mb^2 + nc^2 - mn^2a^2 - nm^2a^2
-> AX^2 = mb^2 + nc^2 - mna^2(n+m) = mb^2 + nc^2 - mna^2, as required.

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