MathDB
Problems
Contests
International Contests
Tournament Of Towns
1995 Tournament Of Towns
(469) 3
(469) 3
Part of
1995 Tournament Of Towns
Problems
(1)
<PAQ = 90^o -/2 <BAC if AP = PK, AQ = QK, 3 angle bisectors
Source: TOT 469 1995 Spring J A3 - Tournament of Towns
7/9/2024
Let
A
K
AK
A
K
,
B
L
BL
B
L
and
C
M
CM
CM
be the angle bisectors of a triangle
A
B
C
ABC
A
BC
, with
K
K
K
on
B
C
BC
BC
. Let
P
P
P
and
Q
Q
Q
be the points on the lines
B
L
BL
B
L
and
C
M
CM
CM
respectively such that
A
P
=
P
K
AP = PK
A
P
=
P
K
and
A
Q
=
Q
K
AQ = QK
A
Q
=
Q
K
. Prove that
∠
P
A
Q
=
9
0
o
−
1
2
∠
B
A
C
.
\angle PAQ = 90^o -\frac12 \angle B AC.
∠
P
A
Q
=
9
0
o
−
2
1
∠
B
A
C
.
(I Sharygin)
geometry
angles
angle bisector