Question

The base of an isosceles triangle is of length $2a$ and if $p$ be its altitude then what is the distance of the mid-point of the base from either of equal sides ?

Answer 1

Choosing the base along $x-$axis and its mid-point as origin so that $B$ is $(-a,0)$ and $C(a,0)$
Now the triangle being isosceles its altitude will be along the median $AO=p$ and will be along $y-$axis. Vertex $A$ be $(0,p)$. By intercepts from the equations of $AB$ and $AC$ are
$\frac{x}{-a}+\frac{y}{p}=1$ and $\frac{x}{a}+\frac{y}{p}=1$
Distance of each from the mid-point $(0,0)$ of base is
$\frac{1}{\sqrt{(\frac{1}{a^2}+\frac{1}{p^2})}}=\frac{ap}{\sqrt{a^2+p^2}}$