Prove that $BD.DC=\frac{AB.AC^2}{2(AB+AC)}$

Question

In $\Delta ABC$, M is the Mid Point of $AC$ and $D$ is a Point on $BC$ such that $DB=DM$ and also $$2(BC)^2-(AC)^2=AB.AC$$

Prove that $$BD.DC=\frac{AB.AC^2}{2(AB+AC)}$$

My try:

Let $BC=a,CA=b,AB=c$

Let $BD=x=DM$

Given that $$2a^2=b(b+c)$$

Also by Appolonius Theorem we have

$$a^2+c^2=\frac{b^2}{2}+2(BM)^2$$

Assuming

$$\angle MBD=\angle BMD=\alpha$$

$$\cos(180-\alpha)=\frac{x^2+x^2-BM^2}{2x^2}$$

Hence we get $$BM=x\cos\left(\frac{\alpha}{2}\right)$$

Any clue here?

Answer 1

By the law of cosines for $\Delta CMD$ we obtain: $$x^2=(a-x)^2+\frac{b^2}{4}-b(a-x)\cdot\frac{a^2+b^2-c^2}{2ab},$$ which gives $$x=\frac{(2a^2-b^2+2c^2)a}{2(3a^2-b^2+c^2)}=\frac{(bc+2c^2)a}{3b^2+3bc-2b^2+2c^2}=\frac{ac}{b+c}$$ and $$DC=a-x=\frac{ab}{b+c}.$$ Id est, $$BD\cdot DC=\frac{a^2bc}{(b+c)^2}=\frac{(b^2+bc)bc}{2(b+c)^2}=\frac{b^2c}{2(b+c)}$$ and we are done!

Answer 2

I have found the following two equations: $$BM^2=\frac{a^2+c^2}{2}-\frac{b^2}{4}$$ and to calculate $BD$ $$BM^2=2BD^2-BD\cdot BM$$ and the third is given by $$2a^2-b^2=b\cdot c$$