在 △ A B C \triangle ABC △ABC 中, D D D, E E E 在 B C BC BC 上 ( D (D (D 靠近 B ) B) B), A D = A E AD=AE AD=AE, 作以 A A A 为圆心, A D AD AD 为半径的圆, 交 ( A B C ) (ABC) (ABC) 于 F F F, G ( F G (F G(F 在不含 C C C 的弧 A B AB AB 上), ( B D K ) (BDK) (BDK) 交 A B AB AB 于 B B B, K K K, ( C E G ) (CEG) (CEG) 交 A C AC AC 于 L L L. 求证: F K FK FK 交 G L GL GL 于 O A OA OA 上.
证明:
显然,
O
A
OA
OA 垂直平分
F
G
FG
FG, 只需证明
∠
K
F
G
=
∠
L
G
F
\angle KFG = \angle LGF
∠KFG=∠LGF.
设 A B AB AB 交 F G FG FG 于 M M M, 设 A C AC AC 交 F G FG FG 于 N N N.
∠ K F G = ∠ A M N − ∠ F K B \angle KFG=\angle AMN-\angle FKB ∠KFG=∠AMN−∠FKB. ∠ L G F = ∠ A N M − ∠ G L C \angle LGF=\angle ANM-\angle GLC ∠LGF=∠ANM−∠GLC.
∠ A M N = ∠ F A B + ∠ A F G \angle AMN=\angle FAB+\angle AFG ∠AMN=∠FAB+∠AFG.
∠ A N M = ∠ G A C + ∠ A G F \angle ANM=\angle GAC+\angle AGF ∠ANM=∠GAC+∠AGF.
∠ A M N − ∠ A N M = ∠ F A B − ∠ G A C \angle AMN-\angle ANM=\angle FAB-\angle GAC ∠AMN−∠ANM=∠FAB−∠GAC.
∠ F K B − ∠ G L C = ∠ F D B − ∠ G E C = ∠ F A E 2 − ∠ D A G 2 = ∠ F A D − ∠ E A G 2 \angle FKB-\angle GLC=\angle FDB-\angle GEC=\frac{\angle FAE}{2}-\frac{\angle DAG}{2}=\frac{\angle FAD-\angle EAG}{2} ∠FKB−∠GLC=∠FDB−∠GEC=2∠FAE−2∠DAG=2∠FAD−∠EAG.
∠ F A D − ∠ E A G = ( ∠ F A B − ∠ G A C ) + ( ∠ B A D − ∠ E A C ) \angle FAD-\angle EAG=(\angle FAB-\angle GAC)+(\angle BAD-\angle EAC) ∠FAD−∠EAG=(∠FAB−∠GAC)+(∠BAD−∠EAC).
∠ B A D − ∠ E A C = ( ∠ A D E − ∠ A B C ) − ( ∠ A E D − ∠ A C B ) = ∠ A C B − ∠ A B C = ( ∠ A G F + ∠ F A B ) − ( ∠ A F G + ∠ G A C ) = ∠ F A B − ∠ G A C \angle BAD-\angle EAC=(\angle ADE-\angle ABC)-(\angle AED-\angle ACB)=\angle ACB-\angle ABC=(\angle AGF+\angle FAB)-(\angle AFG+\angle GAC)=\angle FAB-\angle GAC ∠BAD−∠EAC=(∠ADE−∠ABC)−(∠AED−∠ACB)=∠ACB−∠ABC=(∠AGF+∠FAB)−(∠AFG+∠GAC)=∠FAB−∠GAC.
∠ F A D − ∠ E A G = 2 ( ∠ F A B − ∠ G A C ) \angle FAD-\angle EAG=2(\angle FAB-\angle GAC) ∠FAD−∠EAG=2(∠FAB−∠GAC).
∠ F K B − ∠ G L C = ∠ F A D − ∠ E A G 2 = ∠ F A B − ∠ G A C \angle FKB-\angle GLC=\frac{\angle FAD-\angle EAG}{2}=\angle FAB-\angle GAC ∠FKB−∠GLC=2∠FAD−∠EAG=∠FAB−∠GAC.
∴ \therefore ∴ ∠ A M N − ∠ A N M = ∠ F K B − ∠ G L C \angle AMN-\angle ANM=\angle FKB-\angle GLC ∠AMN−∠ANM=∠FKB−∠GLC, 进而 ∠ K F G = ∠ L G F \angle KFG=\angle LGF ∠KFG=∠LGF.
证毕.
2025年1月11日
扫一扫
被折叠的 条评论
为什么被折叠?
到【灌水乐园】发言
