Question

In the figure, seg $BD\perp sideAC$, $segDE\perp sideBC$, then show that $DE\times BD=DC\times BE$.

Answer 1

In $△BED$ and $△BDC$

$\angle BED=\angle BDC$

$\angle DBE=\angle DBC$

So, $△BED$ and $△BDC$ are similar

If two triangles are similar then there corresponding sides are proportional

$\frac{DE}{DC}=\frac{BE}{BD}$

$⟹$ $DE\times BD=DC\times BE$

Hence proved.