euclidean-geometrygeometryplane-geometry
In In a right triangle $ABC$ ($A=90°$) with inradio $r$, cevian $AD$ is drawn in such a way that the inradium of $ABD$ and $ADC$ are equal to $r1$.If $AD=2$, calculate $\frac{1}{r1}-\frac{1}{r}$ (Answer:0,5).
My progress:
$\triangle CED \sim \triangle CAB \\
\frac{CE}{AC}=\frac{DE}{AB}=\frac{CD}{BC}\\
\triangle BDL \sim \triangle BCA\\
\frac{DL}{AC}=\frac{BD}{BC}=\frac{LB}{AB}\\
CE = CI\\
BK = BL$
but I still haven't found the necessary relationship to finalize
Draw altitude from $B$ to $AC$ and let the foot be $D$. Say $DF=DB=x$. Now applying Pythagorean theorem to $\triangle BCD$ $$(18+x)^2+x^2=(15\sqrt 2)^2\implies x=3$$ So $BD=3$ and $AD=7-3=4$. It can be easily seen that $\triangle BDA$ is a $3:4:5$ triangle and hence $\theta\approx 37^\circ$
$15^\circ$ is not the correct answer. GeoGebra shows the same.
In $\triangle ABD$, traversal $EC$ intersects $BA$ and $BD$ internally and $AD$ externally. Applying Menelaus's theorem,
$ \displaystyle \frac{AC}{CD} \cdot \frac{DN}{NB} \cdot \frac{BE}{EA} = 1 \implies EA = 3 BE$
Or, $EM + MA = 3 BM - 3 EM$.
As $BM = MA$, $BM =2 EM \implies E$ is midpoint of $BM$.
As $CE$ is also the angle bisector, $BC = CM = \sqrt3$ and $CE \perp AB$
If $BE = x$, applying Pythagoras in $\triangle AEC$ and $\triangle BEC$,
$CE^2 = BC^2 - x^2 = AC^2 - (3x)^2 \implies 3 - x^2 = 9 - 9 x^2$
$ \displaystyle x = \frac{\sqrt3}{2}, AB = 2 \sqrt3$
As $BC = CM = MB$, $\triangle BCM$ is equilateral triangle and $\angle ABC = 60^\circ$.
Now given $AD:DC = BA:BC$, $BD$ is angle bisector of $\angle ABC$.
$\therefore \angle ABD = 30^\circ$
(As a side note, $\triangle ACB$ is right triangle with $\angle ACB = 90^\circ$)
Best Answer
Here's what seems to be an unnecessarily-complicated solution.
Define $b:=|AC|$, $c:=|AB|$, $d:=|AD|$, $p:=|BD|$, $q:=|CD|$. Let $r$ be the inradius of $\triangle ABC$, and let $s$ be the common inradius of $\triangle ABD$ and $\triangle ACD$.
We know $$\text{inradius}\cdot \text{perimeter} = 2\,\text{area}$$ so we can write $$\begin{align} s(c+d+p) &= 2|\triangle ABD|=\frac{p}{p+q}\cdot 2|\triangle ABC| = \frac{p}{p+q}\, r (b+c+p+q) \tag1\\[8pt] s(b+d+q) &= 2|\triangle ACD|=\frac{q}{p+q}\cdot 2|\triangle ABC| = \frac{q}{p+q}\, r (b+c+p+q) \tag2 \end{align}$$ Solving this linear system for $b$ and $c$ gives $$ b = -q-d + \frac{2 d q r}{(p + q)(r - s)} \qquad\qquad c = -p-d + \frac{2 d p r}{(p + q)(r - s)} \tag3 $$
Since $\triangle ABC$ is right, we also know $$\begin{align} 2r = |AC|+|AB|-|BC| &= b+c-(p+q) \\ &= 2\,\frac{ ds- (p+q)(r-s)}{r - s} \\ \to \qquad (p+q)(r-s) &= ds -r(r-s)\tag4 \end{align}$$ By Stewart's Theorem, we have $$b^2p+c^2q=(p+q)(d^2+pq) \quad\underset{(3)}{\to}\quad (p + q) s (r - s)= d r (2s-r) \tag5$$
Combining $(4)$ and $(5)$ to eliminate $p+q$ gives
Substituting $d=2$ gives the specific result for the question as stated. $\square$
There's almost-certainly a quicker path to the target relation. Note that $$s(b+c+2d+p+q)=2|\triangle ABC| = r(b+c+p+q) \qquad\to\qquad \frac1s-\frac1r=\frac{d}{|\triangle ABC|}$$ So, really, "all we have to do" is show $d^2=|\triangle ABC|$. I'm not seeing a particularly good way to do that. Even so, looking at this as $2d^2=bc$ gives an easy way to construct an accurate figure in, say, GeoGebra, for further investigation.