ABC is a triangle. D is the center of BC . AC is perpendicular to AD. prove that $\cos(A)\cdot \cos(C)=\frac{2(c^2-a^2)}{3ac}$

Question

ABC is a triangle. D is the center of BC . AC is perpendicular to AD. prove that $$\cos(A)\cdot \cos(C)=\frac{2(c^2-a^2)}{3ac}$$ problem and my attempts are shown in images. I cannot find the exact way to the answer.

Answer 1

so we have to prove that $$\cos(\alpha)\cos(\gamma)=\frac{2(c^2-a^2)}{3ac}$$ we have $$\cos(\gamma)=\frac{2b}{a}$$ (I) and $$\frac{\sin\left(\alpha-\frac{\pi}{2}\right)}{\sin\left(\frac{\pi}{2}+\gamma\right)}=\frac{-cos(\alpha)}{\cos(\gamma)}=\frac{a}{2c}$$ further we have $$b^2+AD^2=\frac{a^2}{4}$$(II) and $$AD^2=\frac{c^2}{2}+\frac{b^2}{2}-\frac{a^2}{4}$$ (III) therefore we get $$a^2-c^2=3b^2$$ and $$\cos(\alpha)\cos(\gamma)=-\frac{2b^2}{ac}=\frac{2(c^2-a^2)}{3ac}$$ if $$3b^2=a^2-c^2$$

Answer 2

Since $cosC=\frac{2b}{a}$, and $cosA=\frac{-b}{c}$, then $$cosA*cosC=\frac{-2b^2}{ac}$$Further,$$a^2=b^2+c^2-2 bc*cosA$$ Substituting $\frac{-b}{c}$ for $cosA$, and transposing gives$$3b^2=a^2-c^2$$From this we get$$-3b^2=c^2-a^2$$or$$\frac{-2b^2}{ac}=\frac{2(c^2-a^2)}{3ac}=cosA*cosC$$