Let ABC be a right-angled triangle with ∠B = 90◦.Let I be the incentre of ABC. Draw a line perpendicular to AI at I. Let it intersect the line CB at D. Prove that CI is perpendicular to AD and prove that ID =sqrt(b(b − a)) ,where BC = a and CA = b.
Answers
In triangle ABC, segment BD is perpendicular to the side AC and segment DE is perpendicular to side BC.
In ΔBDC and ΔBDE, we have
∠1=∠2 (90°)
∠DBC=∠DBE(Common)
Therefore, by AA similarity,
ΔBDC ≅ΔBDE,
By similarity of triangles,
⇒×
=
×
Hence proved.
Answer:
Answer
Solution 1:
We begin with the following lemma:
Lemma: Let XYZ be a triangle with ∠XYZ=90+α. Construct an isosceles triangle XEZ, externally on the side XZ, with base angle.
Then E is the circumcentre of ΔXYZ.
Proof of the Lemma: Draw ED ⊥XZ.
Then DE is the perpendicular bisector of XZ. We also observe that ∠XED=ZED=90−α .
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 ΔXYZ. Join XF.
Then ∠XFD=90−α.
But we know that ∠XED=90−α. 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 ∠PDQ=90
o
. Hence
PQ
2
=PD
2
+DQ
2
=2r
1
2
+2r
2
2
:
Let s
1
= (AB + BD + DA)=2. Observe that BD = ca=b and AD =
AB
2
−BD
2
=
c
2
−(
b
ca
)
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 PQ
2
=2r
2
(
b
2
c
2
+a
2
)=2r
2
Consider ΔPIQ. Observe thatPIQ=90+(B=2)=135.
Hence PQsubtends 90
o
on the circumference of the circumcircle of ΔPIQ. But we have seen that ∠PDQ=90
o
.
Now construct a circle with PQ asdiameter.
Let it cut AC again in K.It follows that ∠PKQ = 90 and thepoints P, D, K, Q are concyclic.
We also notice ∠KPQ=∠KDQ=45
o
and∠PQK=∠PDA=45
o
.
Thus PKQ is an isosceles right-angled triangle with KP = KQ.
∴KP
2
+KQ
2
=PQ
2
=2r
2
and hence KP = KQ = r.
Now ∠PIQ=90+45and∠PKQ=245
o
=90
o
with KP = KQ = r.
Hence K is the circumcentre of ΔPIQ.
(Incidentally,
This also shows that KI = r and hence K is the point of contact of the incircle of ΔABC 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 ΔPIQ.
We show that KP = KQ = r.
Let r
1
andr
2
be the in radii of triangles ABD and CBD respectively.
Draw PL⊥ACandQM⊥AC.
If s
1
isthe semiperimeter of ΔABD, thenAL=s
1
BD.
But s
1
=
2
AB+BD+DA
,BD=
b
ca
,AD=
b
c
2
Hence s
1
= cs=b.
This gives r
1
=s
1
−c=cr/b,AL=s
1
BD=c(sa)=b.
Hence KL=AKAL=(s−a)−
b
c(s−a)
=
b
(b−c)(s−a)
.
We observe that 2r
2
=
2
(c+ab)
2
=
2
c
2
+a
2
+b
2
2bc2ab+2ca
=(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, KP
2
=KL
2
+LP
2
=
b
2
r
2
a
2
+
b
2
r
2
+c
2
=r
2
.
Thus KP = r.
Similarly, KQ = r.
This gives KP = KI = KQ = r and therefore K is the circumcentre of ΔKIQ.
(Incidentally, this also shows that KL=ca=b= r
2
and KM = r
1
.)