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42
Prove CAP=2CPA if BC=AC+0.5AB and BP:PA=1:3
Prove CAP=2CPA if BC=AC+0.5AB and BP:PA=1:3
Source:
October 13, 2010
geometry
circumcircle
geometry unsolved
Problem Statement
Triangle
A
B
C
ABC
A
BC
is given for which
B
C
=
A
C
+
1
2
A
B
BC = AC + \frac{1}{2}AB
BC
=
A
C
+
2
1
A
B
. The point
P
P
P
divides
A
B
AB
A
B
such that
B
P
:
P
A
=
1
:
3
BP : PA = 1 : 3
BP
:
P
A
=
1
:
3
. Prove that
∠
C
A
P
=
2
∠
C
P
A
\angle CAP = 2\angle CPA
∠
C
A
P
=
2∠
CP
A
.
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