Question

$AB$ and $AC$ arc two chords of a circle of radius $r$ such that $AB=2AC$. If $P$ and $Q$ are respectively distance from center to $AB$ and $AC$ then, prove that $4q^2=p^2+3r^2$

Answer 1

Given : From the center $O$ of circle with center $r,p$ and $q$ arc respectively distance to $AB$ and $AC$ such that $AB=2AC$
To Prove : $4q^2=p^2+3r^2$
Construction : Draw perpendicular from center $O$ to chords $AB$ and $AC$ which cuts chord at $M$ and $N$ respectively.

$OM\perp AB$ and $ON\perp AC$

$\therefore AM=MB=\frac{1}{2}AB$ (Let $AB=2x$)

$\therefore AM=MB=\frac{1}{2}\times 2x$

$AM=MB=x$

Similarly,

$ON\perp AC$

$AN=NC=\frac{1}{2}AC$

$AN=NC=\frac{1}{2}\times \frac{1}{2}AB[\because AB=2AC,\therefore AC=\frac{1}{2}AB]$

$=\frac{1}{2}\times \frac{1}{2}\times 2x$

$=\frac{x}{2}$

$AN=NC=\frac{x}{2}$

In right angles $\Delta AOM$

$O{A}^2=A{M}^2+M{O}^2$

$r^2=x^2+p^2$

$x^2=r^2-p^2$................(i)

In right angled $\Delta AON$

$A{O}^2=A{N}^2+O{N}^2$

$r^2=(\frac{x}{2}{)}^2+q^2$

$(\frac{x}{2}{)}^2=r^2-q^2$

$\frac{x^2}{4}=r^2-q^2$

$x^2=4(r^2-q^2)$............(ii)

From equation (i) and (ii)

$r^2-p^2=4(r^2-q^2)$

$r^2-p^2=4r^2-4q^2$

$4q^2=4r^2-r^2-r^2+p^2$

$4q^2=p^2+3r^2$