If the base of an Isosceles triangle is of length ' 2a ' and length of altitude dropped to the base is ' h ' then the distance from mid point of the base to the side of triangle

Let ‘p’ be the distance from the midpoint of the base to the equal side of the triangle

$$\begin{gathered} given AD=h \\ D is the midpoint of BC then DC=a \\ In ∆ADC AC=\sqrt{A{D}^2+D{C}^2} \\ =\sqrt{h^2+a^2} \\ area of ∆ABC=2 [area of ∆ADC] \end{gathered}$$

$\frac{1}{\cancel{2}}\times \cancel{2}a\times h=\cancel{2}\left[\frac{1}{\cancel{2}}\times p\times \sqrt{h^2+a^2}\right]$

 $\Rightarrow ah=p.\sqrt{h^2+a^2}$

$\Rightarrow p=\frac{ah}{\sqrt{h^2+a^2}}$