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University of Regina
Department of Mathematics and Statistics
MATH231 – Euclidean Geometry – Winter 2017
Homework Assignment No. 5 - Solutions
1. The quadrilateral BV P U is cyclic (∠BV P = π2 , so V is on the circle of diameter BP ;
also, since ∠BU P = π2 , U is on the circle of diameter BP ; so both U and V are on
the circle of diameter BP ). Similarly one shows that the quadrilateral CW V P are
cyclic. In the circle circumscribed to BV P U , the angles ∠BV U and ∠BP U subtend
the same circular arc, so they are equal. We showed
∠BV U = ∠BP U.
In the circle circumscribed to CW V P , the angles ∠W V C and ∠W P C subtend the
same circular arc, so they are equal. We showed
∠W V C = ∠W P C.
It remains to show that ∠BP U = ∠W P C. To this end, it is sufficient to show that
∠U P W = ∠BP C. Indeed, the quadrilateral AU P W is cyclic (because the sum of
the opposite angles AU P and AW P is equal to π/2 + π/2 = π). We deduce that
∠U P W + ∠U AW = π ⇒ ∠U P W = π − ∠U AW.
Also the quadrilateral ABP C is cyclic (because P has been chosen on the circumcircle
of ABC). We deduce that
∠BP C + ∠BAC = π ⇒ ∠BP C = π − ∠BAC.
So ∠U P W = ∠BP C, as desired.
In conclusion, we have proved that ∠BV U = ∠CV W . This implies that the points
U, V , and W are collinear.
2*. As the hint says, let us show that ∠AA1 B1 = ∠AA1 C1 . To this end, we first note that
the quadrilateral HA1 BC1 is cyclic, because ∠HA1 B + ∠HC1 B = π/2 + π/2 = π.
On the circle circumscribed to this quadrilateral, the angles ∠HA1 C1 and ∠HBC1
subtend the same circular arc, so they are equal. We note this:
∠HA1 C1 = ∠HBC1 .
The quadrilateral HA1 CB1 is cyclic as well, because ∠HA1 C + ∠HB1 C = π/2 +
π/2 = π. On the circle circumscribed to this quadrilateral, the angles ∠HA1 B1 and
∠HCB1 subtend the same circular arc, so they are equal. We note this:
∠HA1 B1 = ∠HCB1 .
We finally show that ∠HBC1 = ∠HCB1 . Indeed, the triangle ABB1 is right angled,
so
∠HBC1 = ∠ABB1 = π/2 − ∠BAC.
Also the triangle ACC1 is right angled, so
∠HCB1 = ∠C1 CA = π/2 − ∠BAC.
We see that ∠HBC1 = ∠HCB1 , as desired.
Similarly, we show that ∠C1 B1 B = ∠A1 B1 B and ∠A1 C1 C = ∠B1 C1 C. This
means that AA1 , BB1 , and CC1 are the bisectors of the triangle A1 B1 C1 . So their
intersection H is the centre of the incircle of the triangle A1 B1 C1 .
1
2
3*. We follow the hint. First, ∆AA′ Q ∼ ∆CP Q, because ∠AA′ Q = ∠QP C and
∠A′ AQ = ∠QCP (as alternate angles). We deduce
CQ
CP
.
=
QA
AA′
Then, ∆AA′ R ∼ ∆BP R, because ∠AA′ R = ∠RP B and ∠A′ AR = ∠RBP (as
alternate angles). We deduce
AA′
AR
=
.
RB
BP
From the previous two equations we deduce
BP CQ AR
BP CP AA′
·
·
·
=
·
= 1.
P C QA RB
P C AA′ BP
4*. See Figure 3. In the triangle BHC, B ′ is the midpoint of the side BH and C ′ the
midpoint of the side CH. We deduce that B ′ C ′ ∥BC and
1
B ′ C ′ = BC.
2
Similarly we prove that
1
1
A′ B ′ = AB, A′ C ′ = AC.
2
2
This implies
A′ B ′
B′C ′
A′ C ′
1
=
=
= .
AB
BC
AC
2
This implies the desired conclusion: ∆A′ B ′ C ′ ∼ ∆ABC, the proportionality factor
being 1/2.
A
B1
C1
H
B
A1
Figure 2
C
3
A
A'
H
B'
C'
B
C
Figure 3