Площа дельтоїда через діагоналі

Suppose we take a kite $ABCD$ such that $AB=BC$ and $CD=DA$. Suppose that the diagonals $AC$ and $BD$ intersect at $X$.

By SSS congruence, $\triangle{ABD}\cong\triangle{CBD}$. Thus, $\angle{ABD}\cong\angle{CBD}$. By SAS congruence, $\triangle{ABX}\cong\triangle{CBX}$. This means that $\angle{AXB}\cong\angle{CXB}$, which means that $AC$ and $BD$ are perpendicular.

Since $AC\perp BD$, $ABX$, $BCX$, $CXD$, and $DAX$ are all right triangles.

Computing their areas, $$[ABX]=\frac{1}{2}XA\cdot XB$$ $$[BCX]=\frac{1}{2}XB\cdot XC$$ $$[CDX]=\frac{1}{2}XC\cdot XD$$ $$[DAX]=\frac{1}{2}XD\cdot XA$$

Computing $[ABCD]$, $$[ABCD]=[ABX]+[BCX]+[CDX]+[DAX]$$ $$[ABCD]=\frac{1}{2}(XA\cdot XB+XB\cdot XC+XC\cdot XD+XD\cdot XA)$$ $$[ABCD]=\frac{1}{2}(XA+XC)(XB+XD)$$ $$[ABCD]=\frac{1}{2}AC\cdot BD$$

Thus, the area of a kite is equal to one half of the product of the diagonals.

In fact, this formula applies to any quadrilateral with perpendicular diagonals.