Question

球一到数学题P是三角形ABC内一点,连接AP,BP,CP并延长,分别于BC,AC,AB交与D,E,F点,已知：AP=6,

Answer 1

过A作BC的平行线分别交BE、CF的延长线于G、H

AP=PD ==>AG=BD,BP=PG=9 ==>EG=9-3=6

==>AG：BC=EG：BE=6：12=1/2 ==>D为BC中点

同理：AH=CD=BC/2 ==>FH=CF/2=10 ==>CP=(10+20)/2=15

又AG‖=CD ==>CG=AD=12

△CPG中,PG²+CG²=9²+12²=15²=CP² ==>∠CGP是直角

==>S△ABC=S△GBC=CG•BG/2=12•18/2=108