Question

Let $ABC$ be a right-angled triangle with $\angle B=90.$ Let $BD$ be the altitude from $B$ on to $AC.$ Let $P,Q$ and $I$ be the incentres of triangles $ABD,CBD,$ and $ABC,$ respectively. Show that the circumcentre of the triangle $PIQ$ lies on the hypotenuse $AC.$

Answer 1

Solution 1:
We begin with the following lemma:
Lemma: Let XYZ be a triangle with $\angle XYZ=90+\alpha$. Construct an isosceles triangle XEZ, externally on the side XZ, with base angle.
Then E is the circumcentre of $\Delta$XYZ.
Proof of the Lemma: Draw ED $\perp$XZ.
Then DE is the perpendicular bisector of XZ. We also observe that $\angle XED=ZED=90-\alpha$ .
Observe that E is on the perpendicular bisector of XZ.
Construct the circumcircle of XY Z. Draw perpendicular bisector of XY and let it meet DE in F.
Then F is thecircumcentre of $\Delta$XYZ. Join XF.
Then $\angle XFD=90-\alpha$.
But we know that $\angle XED=90-\alpha$. Hence E = F.
Let $r_1,r_2$ and r be the inradii of the triangles ABD, CBD and ABCrespectively. Join PD and DQ. Observe that $\angle PDQ={90}^o$. Hence
$P{Q}^2=P{D}^2+D{Q}^2=2r_1^2+2r_2^2$:
Let $s_1$ = (AB + BD + DA)=2. Observe that BD = ca=b and AD = $\sqrt{A{B}^2-B{D}^2}=\sqrt{c^2-{\left(\frac{ca}{b}\right)}^2}=c^2/b$.
This gives $s_1$ = cs=b. But $r_1=s_1$ c = (c=b)(s b) = cr=b. Similarly, $r_2$ = ar=b.
Hence $P{Q}^2=2r^2\left(\frac{c^2+a^2}{b^2}\right)=2r^2$
Consider $\Delta$PIQ. Observe that$PIQ=90+(B=2)=135$.
Hence PQsubtends ${90}^o$ on the circumference of the circumcircle of $\Delta PIQ$. But we have seen that $\angle PDQ={90}^o$.
Now construct a circle with PQ asdiameter.
Let it cut AC again in K.It follows that $\angle PKQ$ = 90 and thepoints P, D, K, Q are concyclic.
We also notice $\angle KPQ=\angle KDQ={45}^{o}and\angle PQK=\angle PDA={45}^o$.
Thus PKQ is an isosceles right-angled triangle with KP = KQ.
$\therefore K{P}^2+K{Q}^2=P{Q}^2=2r^2$ and hence KP = KQ = r.
Now $\angle PIQ=90+45and\angle PKQ=2{45}^o={90}^o$ with KP = KQ = r.
Hence K is the circumcentre of $\Delta PIQ$.
(Incidentally,
This also shows that KI = r and hence K is the point of contact of the incircle of $\Delta$ABC with AC.)
Solution 2:
Here we use computation to prove that the point of contact K of the incircle with AC is the circumcentre of $\Delta PIQ$.
We show that KP = KQ = r.
Let $r_{1}and{r}_2$ be the in radii of triangles ABD and CBD respectively.
Draw $PL\perp ACandQM\perp AC$.
If $s_1$ isthe semiperimeter of $\Delta ABD$, then$AL=s_{1}BD$.
But $s_1=\frac{AB+BD+DA}{2},BD=\frac{ca}{b},AD=\frac{c^2}{b}$
Hence $s_1$ = cs=b.
This gives $r_1=s_1-c=cr/b,AL=s_{1}BD=c\left(sa\right)=b$.
Hence $KL=AKAL=(s-a)-\frac{c(s-a)}{b}=\frac{(b-c)(s-a)}{b}$.
We observe that $2r^2=\frac{(c+ab{)}^2}{2}=\frac{c^2+a^2+b^{2}2bc2ab+2ca}{2}=(b^{2}ba-bc+ac)=(b-c)(b-a)$.
This gives
(s - a)(b- c) = (s -b + b -a)(b- c) = r(b -c) + (b -a)(b- c)
= r(b-c) + $2r^2$ = r(b-c + c + a-b) = ra.
Thus KL = ra=b.
Finally, $K{P}^2=K{L}^2+L{P}^2=\frac{r^2{a}^2}{b^2}+\frac{r^2+c^2}{b^2}=r^2$.
Thus KP = r.
Similarly, KQ = r.
This gives KP = KI = KQ = r and therefore K is the circumcentre of $\Delta KIQ$.
(Incidentally, this also shows that $KL=ca=b=$ $r_2$ and KM = $r_1$.)