$a,b,c,e$ and $a,c,d,e$ form a concave quadrilateral then $a,b,d,e$ form a concave quadrilateral?

Question

Let $a,b,c,d,e$ be five lines on the plane with distinct slopes,

If $a,b,c,e$ form a concave quadrilateral and $a,c,d,e$ form a concave quadrilateral, does it imply that $a,b,d,e$ form a concave quadrilateral?

To check whether four lines form a concave quadrilateral, it is enough to check their slopes satisfy the inequality in The concave quadrilateral and the slopes of its sides

So the problem becomes an inequality problem:

Let $a,b,c,d,e\in\mathbb{R}$ be distinct, \begin{align*}(a-b) (e-a) (b-c) (c-e)<0,\\(a-c) (e-a) (c-d) (d-e)<0,\\\text{Prove：}(a-b) (e-a) (b-d) (d-e)<0.\end{align*}

Answer 1

WolframAlpha checks $$\frac{(a-e) (b-d)}{(a-b) (e-d)}=1-\left(1-\frac{(e-a) (b-c)}{(a-b) (c-e)}\right) \left(1-\frac{(e-a) (c-d)}{(a-c) (d-e)}\right)$$is an identity.

The problem supposes $\frac{(e-a) (b-c)}{(a-b) (c-e)}<0$ and $\frac{(e-a) (c-d)}{(a-c) (d-e)}<0$, then $1-\frac{(e-a) (b-c)}{(a-b) (c-e)}>1$ and $1-\frac{(e-a) (c-d)}{(a-c) (d-e)}>1$, therefore the right hand side of the identity is $<0$, therefore $\frac{(a-e) (b-d)}{(a-b) (e-d)}<0$.