Interpreting a geometry problem involving chords of a circle

Question

The problem is stated as follows:

Prove that if from the endpoints of a diameter of a circle, two intersecting chords are drawn, then the sum of the products of each chord and the segment of it from the endpoint of the diameter to the intersection point is a constant quantity.

I do not want to do the problem if I did not interpret it correctly. Below is an image based on what I understand. If the figure is correct, am I to prove that $\vert AX\vert \vert AN\vert + \vert BX\vert \vert BM\vert=$ constant?

Answer 1

Let $\measuredangle A=\alpha$, $\measuredangle B=\beta$ and $AB=2R$.

Thus, by law of sines for $\Delta ABX$ we obtain:

$$AX=\frac{2R\sin\beta}{\sin(\alpha+\beta)}$$ and

$$BX=\frac{2R\sin\alpha}{\sin(\alpha+\beta)}.$$

Also since $\measuredangle AMB=\measuredangle ANB=90^{\circ},$ we obtain: $$AN=2R\cos\alpha$$ and $$BM=2R\cos\beta.$$ Thus, $$AX\cdot AN+BX\cdot BM=\frac{2R\sin\beta\cdot2R\cos\alpha+2R\sin\alpha\cdot2R\cos\beta}{\sin(\alpha+\beta)}=4R^2.$$ Done!

Answer 2

Hint:  let the centre of the circle be $\,O\,$ and $AB=2r\,$, then:

• $\;\vert AX\vert \vert XN\vert = \vert BX\vert \vert XM\vert= r^2 - |XO|^2\,$ by the power of point X formulas;

• $\;|AX|^2+|BX|^2 = 2 r^2 + 2 |XO|^2\,$ by the triangle median length formula.

Use the above in $\,|AX||AN|+|BX||BM|=|AX|\big(|AX|+|XN|\big)+|BX|\big(|BX|+|XM|\big)\,$.