Topic 1: Menelaus’s Theorem, ceve’s
Theorem and related problems
Group 4: Leader: Giang Giang
PROBLEM 2. Let P be a point in the plane of triangle ABC, and ` a line passing
through P . Let A0 , B 0 , C 0 be the points where the reflections of lines P A, P B, P C with
respect to ` intersect lines BC, AC, AB respectively. Prove that A0 , B 0 , C 0 are collinear.
(Hint: Using the Law of Sines.)
Solution
Figure 1:
Apply law of sines. I can see that:
sin ∠AP B 0
sin ∠AB 0 P
sin ∠CP A0
CA0 = CP ·
sin ∠CA0 P
sin ∠BP C 0
BC 0 = BP ·
sin ∠BC 0 P
AB 0 = AP ·
sin ∠CP B 0
sin ∠CB 0 P
sin ∠BP A0
A0 B = BP ·
sin ∠BA0 P
sin ∠AP C 0
C 0 A = AP ·
.
sin ∠AC 0 P
B 0 C = CP ·
Therefore
AC 0
AP sin ∠AP C 0
=
,
BC 0
BP sin ∠BP C 0
BA0
BP sin ∠BP A0
=
,
CA0
CP sin ∠CP A0
CB 0
CP sin ∠CP B 0
=
.
AB 0
AP sin ∠AP B 0
The construction of A0 , B 0 , C 0 by reflections using axis of symmetry property and property
of two complementary angles.
sin ∠AP B 0 = sin ∠BP A0
(supplementary angles)
sin ∠CP A0 = sin ∠AP C 0
(supplementary angles)
sin ∠BP C 0 = sin ∠CP B 0 .
Hence,
AC 0 BA0 CB 0
·
·
= 1,
BC 0 CA0 AB 0
and the proof is complete by Menelaus’ theorem.
PROBLEM 1. Points A1 , B1 , C1 are chosen on the sides BC, CA, AB, respectively of a
triangle ABC. Denote by Ga , Gb , Gc are the centroids of triangles AB1 C1 , BC1 C1 , CA1 B1 ,
respectively. Prove that the lines AGa , BGb , CGc are concurrent if and only if lines
AA1 , BB1 , CC1 are concurrent.
Solution
Figure 2:
Let G, H, I be a mid point of C1 B1 , B1 A1, A1 C1 .
Using law of sines, when G be a mid point of B1 C1 . we have that
S4AB1 G
AG.AB1 sin ∠GAB1
AB1 sin ∠GAB1
=
=
·
=1
S4AC1 G
AG.AC1 . sin ∠GAC1
AC1 sin ∠GAC1
Then
sin ∠GAB1
AC1
=
sin ∠GAC1
AB1
Similar, we also have
B1 C
sin ∠HCB1
=
;
sin ∠HCA1
A1 C
sin ∠IBA1
A1 B
=
sin ∠IBC1
C1 B
Therefore, apply Ceve’s Theorem for three concurrent line AA1 , BB1 , CC1 :
A1 B B1 C
·
A1 C B1 A
sin ∠GAB1
⇒
sin ∠GAC1
C1 A
=1
C1 B
sin ∠HCB1 sin ∠IBA1
·
·
=1
sin ∠HCA1 sin ∠IBC1
·
Apply Ceve’s Theorem for triangle ABC we have AG, BI, CH be concurrent or know
that Ga A, Gb B, Gc C be median line respect to 4AC1 B1 , 4BC1 A1 , 4CB1 A1 .
Prove the opposite is similar with Ga , Gb , Gc be concurrent, we also have that:
sin ∠GAC1 sin ∠HCB1 sin ∠IBA1
A1 B B1 C C1 A
·
·
=1
·
·
=
sin ∠GAC1 sin ∠HCA1 sin ∠IBC1
A1 C B1 A C1 B
Therefore, we conlude that AA1 , BB1 , CC1 be concurrent follow Ceve’s Theorem.
PROBLEM 3. Let ABC be a triangle. Let AM, BN, CP be its internal angle-bisectors
(M ∈ BC, N ∈ CA, P ∈ AB). Find the measure of angle BAC so that P M is perpendicular to NM.
Solution
Figure 3:
Let we sign that ∠BAC = ∠A, ∠ABC = ∠B, ∠BCA = ∠C
Draw a line parallel to BC, intersecting M P and M N at D and E. Apply Ceve Theorem
for CP , BN , AM , we have that
P A M B CN
·
·
=1
P B M C AN
Other hand, we hnow that follow Thales Theorem have
PA
DA
=
,
PB
BM
CN
MC
=
AN
AE
Therefore
DA M B M C
·
·
=1
BM M C AE
⇒DA = AE
Then consider right triangle 4M DE at M have DA = AE then AM = 12 DE = AD = AE
because AM be median line.
When 4ADM and 4AM E be a isosceles triangle at A. we conclude that ∠ADM =
∠AM D
Other hand ∠ADM = ∠DM B so that DM be a bisector of ∠AM B.
Similarly, we also have M E be a bisector of ∠AM C.
Consider 4ABM have
1
∠AM B = ∠M AC + ∠M CA = ∠A + ∠C
2
1
1
1
∠AM B = ∠P M O = ∠A + ∠C
2
4
2
we have ∠P OM = ∠AOC (two opposite angles) then
∠M P O + ∠OM P = ∠OAC + ∠OCA
1
1
1
1
⇔ ∠M P O + ∠A + ∠C = ∠A + ∠C
4
2
2
2
1
⇔ ∠M P O = ∠A
4
Similarly proof we also have ∠M N O = 14 ∠A.
Consider triangle 4ABM have F be intersection of two bisector BO and M P therefore
AF also be a bisector then ∠P AF = ∠OAF = 14 ∠A.
Consider quadrilateral AP F O have ∠M P O = ∠F P O = ∠OAF =
∠F P O and ∠OAF intercept the same art OF .
⇒ Quadrilateral AP F O be a inscribed quadrilateral.
1
∠A.
4
Moreover
The last proof we consider 4BF M have
∠F BM + ∠F M B + ∠BF M = 180◦
1
1
1
⇒ ∠B + ∠A + ∠C + 180◦ − ∠F P O − ∠P OF = 180◦
2
4
2
1
1
1
1
⇒ ∠B + ∠A + ∠C + 180◦ − ∠A = 180◦
2
4
2
2
We additon
1
1
∠A − ∠A to left hand
2
2
1
3
1
1
⇒ ∠B − ∠A + ∠C + ∠A = 0◦
2
4
2
2
3
⇒ ∠A = 90◦
4
⇒∠A = 120◦