In traingle ABC AD is median ,ED and FD are the angle bisectors of angles ADB and ADC respectively,then show that EF || BC.
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In tri. DAE, DE Bisect angle ADB
So we have
DA/DB=AE/EB equ.1
Similarly in tri. DAC, DE Bisect angle ADC
we get
DA/DC=AF/FC
(DC=DB)
DA/DB=AF/FC equ.2
From equ 1 and equ 2
AE/EB= AF/FC
In tri ABC
EF ¶ BE (... BPT)
So we have
DA/DB=AE/EB equ.1
Similarly in tri. DAC, DE Bisect angle ADC
we get
DA/DC=AF/FC
(DC=DB)
DA/DB=AF/FC equ.2
From equ 1 and equ 2
AE/EB= AF/FC
In tri ABC
EF ¶ BE (... BPT)
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