Basis for Euclidean topology on $\mathbb{R}^2$ I am referring book Topology Without Tears
(i)In the book it is said,$\mathcal{B}$ collection of all 'open rectangles' $\{(x,y):(x,y)\in \mathbb{R}^2,a<x<b,c<y<d\}$ in the plane which have sides parallel to the $X$ axis or $Y$ axis is the basis for a topology on plane called Euclidean Topology.
(ii)It is further also proved : $\mathcal{B}$ set of all "open" equilateral traingles with the base parallel to the $X$ axis is a basis for the euclidean topology on $\mathbb{R}^2$
(iii)It is further stated " open rectangle " can be replaced by "open discs"
And there is a theorem for a collection to be basis of any topology:
Let $X$ be a non-empty set and $\mathcal{B}$ be a collection of subsets of $X$.Then $\mathcal{B}$ is a basis for a topology on $X$ if and only if $B$ has following properties:
(a)$X=\cup_{B\in \mathcal{B}}B $
(b)For any $B_1,B_2\in \mathcal{B}$ ,the set $B_1\cap B_2$ is a union of members of $\mathcal{B}$
From the above theorem the collection in (i), can truly be a basis for a topology as plane,$\mathbb{R}^2$ is the union of all of the open rectangle and intersection of two rectangle is a rectangle.
My question is how the collection of equilateral traingle with base parallel to $x$ axis can form basis according to the theorem above?
Because $\mathbb{R}^2$ may be seen as union of these open traingles but the intersection of two equilateral traingle need not be a equilateral traingle? or can it be written as union of equilateral traingle?
Same question in the case of disc,can intersection of two open disc can be written as union of open disc?
 A: By $B(x,r)$ I will denote an open ball around $x$ of radius $r$.
Intersection of two open discs need not be a disc. But it can be written as union of open discs. Take two open discs $D_1$ and $D_2$ and $x\in D_1\cap D_2$. Since $x\in D_1$ then there is $r_1>0$ such that $B(x,r_1)\subseteq D_1$. Analogously there is $r_2>0$ such that $B(x,r_2)\subseteq D_2$. Then for $r(x):=\min(r_1,r_2)$ we have $B(x,r(x))\subseteq D_1\cap D_2$. And so $D_1\cap D_2=\bigcup_{x\in D_1\cap D_2}B(x,r(x))$.
Analogously the intersection of equilateral triangles can be written as a union of equilateral triangles. Regardless of whether their sides are parallel to $x$, $y$ or whatever axis. For every point in an intersection $T_1\cap T_2$ you need to find a small enough equliateral triangle around $x$ contained in $T_1\cap T_2$. It is a bit harder compared to discs, but here's a sketch: first for $x\in T_1\cap T_2$ find an open disc $B(x,r)$ around $x$ fully contained in $T_1\cap T_2$. For that you need a well defined distance $r$ of $x$ from each side of $T_1$ and $T_2$. Then take an equilateral triangle around $x$ inside $B(x,r)$. The union of all such triangles around every point is of course $T_1\cap T_2$.
A: Beside the expressions of @freakish, notice that since your triangles have parallel bases, and therefore other sides too, the intersection of those becomes equilateral one too, also its base is parallel to others.
