Question

AP = x, PC = y and OP = r, the in radius of the right angled triangle ABC, with B right angled.

Find the value of (1 + $\frac{r}{x}$)(1 + $\frac{r}{y}$)

Answer 1

Observe that quadrilateral OQBR has $\angle QBR$ = ${90}^{\circ }$.

Since AB and BC are tangents to the circle, OQ and OR are perpendicular to AB and BC respectively and so $\angle BQO$ and $\angle BRO$ are also ${90}^{\circ }$. Since three angles in a quadrilateral are ${90}^{\circ }$, the fourth angle will be ${360}^{\circ }$ - 3( ${90}^{\circ }$) = ${90}^{\circ }$. So it is a rectangle. Also, since OQ = OR, it follows that all the sides are equal and hence quadrilateral OQBR is actually a square. So, BQ = BR = r.

$\Rightarrow {(x+y)}^2$ = ${(r+x)}^2$ + ${(r+y)}^2$

$\Rightarrow x^2+y^2+2xy=r^2+x^2+2rx+r^2+y^2+2ry$

$\Rightarrow 2xy=2r^2+2rx+2ry$

$\Rightarrow xy=r^2+r(x+y)$

$\Rightarrow r^2+r(x+y)+xy=xy+xy$

$\Rightarrow (r+x)(r+y)=2xy$

$\Rightarrow$ $\frac{(r+x)(r+y)}{xy}$ = 2

$\Rightarrow$ ( $\frac{r+x}{x}$)( $\frac{r+y}{y}$) = 2

$\Rightarrow$(1 + $\frac{r}{x}$)(1 + $\frac{r}{y}$) = 2