Prove that $d^2=a^2+b^2+c^2-2[ab\cos B+bc\cos C+ac\cos(α)]$

Question

The angle between the AB and CD sides of an ABCD convex quadrilateral is equal to $\alpha$. Considering that AB = a, BC = b, CD = c, DA = d, prove that:

$$d^2=a^2+b^2+c^2-2[ab\cos B+bc\cos C+ac \cos(\alpha)]$$ I tried to prove this by Cosines Law, but I couldn't...

Can someone help me? Thanks for antetion.

Let be E=AB∩CD. I applied the cosines law in the triangles AED, ABD and BCE but I couldn't prove this.

Answer 1

We can end your idea by the following way.

Let $AB\cap CD=\{E\}$, $EB=x$ and $EC=y$, $\measuredangle ABC=\beta$ and $\measuredangle BCD=\gamma$.

Thus, $$d^2=(a+x)^2+(c+y)^2-2(a+x)(c+y)\cos\alpha$$ or $$d^2=a^2+c^2-2ac\cos\alpha+x^2+y^2-2xy\cos\alpha+2ax+2cy-2cx\cos\alpha-2ay\cos\alpha$$ and since $$x^2+y^2-2xy\cos\alpha=b^2,$$ it's enough to prove that: $$a(x-y\cos\alpha+b\cos\beta)+c(y-x\cos\alpha+b\cos\gamma)=0$$ or $$a\left(x-\frac{x^2+y^2-b^2}{2x}-\frac{x^2+b^2-y^2}{2x}\right)+c\left(y-\frac{x^2+y^2-b^2}{2y}-\frac{y^2+b^2-x^2}{2y}\right)=0,$$ which is obvious.

Answer 2

Let $AC=d_1$, $BD=d_2$, intersection of $AB$ and $CD$ be $E$, $BE=a_1$,$CE=c_1$, then using cosine law on $\triangle AED$, $\triangle ABC$, $\triangle BCD$, $\triangle BEC$, $\triangle AEC$, and $\triangle DEB$ we get:

\begin{align} d^2&=(a+a_1)^2+(c+c_1)^2-2(a+a_1)(c+c_1)\cos\alpha\\ d_1^2&=a^2+b^2-2ab\cos B\\ d_2^2&=c^2+b^2-2cb\cos C\\ b^2&=a_1^2+c_1^2-2a_1c_1\cos\alpha\\ d_1^2&=(a+a_1)^2+c_1^2-2(a+a_1)c_1\cos\alpha\\ d_2^2&=a_1^2+(c+c_1)^2-2a_1(c+c_1)\cos\alpha \end{align}

Now sum first four equations and minus last two to get what you want.