Question

Let $ABC$ be a right-angled triangle with $\angle B={90}^0$. Let $D$ be a point on $AC$ such that in-radii of the triangle $ABD$ and $CBD$ are equal. If this common value is $r^{\prime }$ and if $r$ is the in-radius of triangle $ABC$, then

• A. $r^{\prime }=r+BD$
• B. $\frac{AC}{r^{\prime }}=1+\frac{r}{BD}$
• C. $\frac{BD}{r^{\prime }}=1+\frac{r}{AC}$
• D. $\frac{1}{r^{\prime }}=\frac{1}{r}+\frac{1}{BD}$

Answer 1

Answer: (D)

$\frac{1}{r^{\prime }}=\frac{1}{r}+\frac{1}{BD}$

Let $E$ and $F$ be the incentres of triangles $ABD$ and $CBD$ respectively.

Let the incircles of triangles $ABD$ and $CBD$ touch $AC$ in $P$ and $Q$ respectively.

If $\angle BDA=\theta$, we see that

$r^{\prime }=PD\tan \frac{\theta }{2}=QD\cot \frac{\theta }{2}$

Hence,

$PQ=PD+DQ=r^{\prime }\left(\cot \frac{\theta }{2}\right)=\frac{2r^{\prime }}{\sin \theta }$

But we observe that,

$DP=\frac{BD+DA-AB}{2},DQ=\frac{BD+DC-BC}{2}$

Thus,

$PQ=\frac{b-c-a+2BD}{2}$. We also have

$\frac{ac}{2}=\left[ABC\right]=\left[ABD\right]+\left[CBD\right]=r^{\prime }\frac{AB+BD+DA}{2}+r^{\prime }\frac{CB+BD+DC}{2}$

$=r^{\prime }\frac{c+a+b+2BD}{2}=r^{\prime }(s+BD)$

But $r^{\prime }=\frac{PQ\sin \theta }{2}=\frac{PQ\cdot h}{2BD}$

where $h$ is the altitude from $B$ on to $AC$.

But we know that $h=\frac{ac}{b}$.

Thus we get

$ac=2\times r^{\prime }(s+BD)=2\times \frac{PQ\cdot h}{2\times BD}(s+BD)=\frac{(b-c-a+2BD)ca(s+BD)}{2\times BD\times b}$

This we get,

$2\times BD\times b=2\times (BD-(s-b\left)\right)(s+BD)$

This gives,

$b^2=s(s-b)$

Since, $ABC$ is a right angled triangle $r=s-b$

Thus we get $B{D}^2=rs$

On the other hand, we also have $\left[ABC\right]=r\text{'}(s+BD)$. Thus we get

$rs=\left[ABC\right]=r\text{'}(s+BD).$

Hence

$\frac{1}{r^{\prime }}=\frac{1}{r}+\frac{BD}{rs}=\frac{1}{r}+\frac{1}{BD}$