AD is a median of triangle ABC. The bisector of angle ADB and angle ADC meet AB and AC in E and F respectively. Prove that EF || BC.
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Answer:
In △DAE, DE Bisect ∠ADB
So we have
DB
DA
=
EB
AE
------1
Similarly in △DAC, DE Bisect ∠ADC
we get
DC
DA
=
FC
AF
--- (DC=DB)
DB
DA
=
FC
AF
------2
From 1 and 2
=
EB
AE
=
FC
AF
In △ABC,
EF∥BE (Baisc Proportionality Theorem)
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