Inm triangle ABC, ADisthemedian. The internal bisector of angle ADB and angle ADC meet AB and AC in E and F respectively. Prove that EF II BC.
Hi!
Here is the answer to your question.
The internal bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angles.
.png)
Given: In ∆ABC, AD is the median.
DE and DF are internal bisectors of ∠ADB and ∠ADC respectively.
To prove: EF||BC
Proof:
In ∆ABD, DE is the bisector of ∠ADB.
.png)
Cheers!