在 △ A B C \triangle ABC △ABC 中, A D AD AD, B E BE BE, C F CF CF 为三条高线, 延长 E F EF EF 交 ( A B C ) (ABC) (ABC) 于 P P P, 延长 D F DF DF, B P BP BP 交于 Q Q Q. 求证: A P = A Q AP=AQ AP=AQ.
证明:
设
D
F
DF
DF 交
(
A
P
F
)
(APF)
(APF) 于点
Q
′
Q'
Q′,
∠
Q
′
F
A
=
∠
Q
′
P
A
=
∠
D
F
B
=
∠
A
C
B
=
π
−
∠
A
P
B
\angle Q'FA=\angle Q'PA=\angle DFB=\angle ACB=\pi-\angle APB
∠Q′FA=∠Q′PA=∠DFB=∠ACB=π−∠APB. 所以
Q
′
Q'
Q′,
P
P
P,
A
A
A 共线,
Q
′
Q'
Q′ 即为
Q
Q
Q.
∠
Q
A
P
=
∠
Q
F
P
=
∠
D
F
E
=
π
−
2
∠
A
C
B
\angle QAP=\angle QFP=\angle DFE=\pi-2\angle ACB
∠QAP=∠QFP=∠DFE=π−2∠ACB, 进而
∠
P
Q
A
=
∠
A
C
B
=
∠
Q
P
A
\angle PQA=\angle ACB=\angle QPA
∠PQA=∠ACB=∠QPA.
Q
A
=
Q
P
QA=QP
QA=QP.
证毕.
