Question

In the following figure, AD is the median of ΔABC. The bisectors of ∠ADB and ∠ADC meet AB and AC in points E and F, respectively. Prove that: EF || ВС.

Answer 1

Given:

In △ABC, AD is the median. The bisectors of ∠ADB and ∠ADC meet AB and AC at E and F, respectively.

Since AD is the median of △ABC,

BD = DC

⇒ Now, in △ADB, DE bisects ∠ADB.

So, by the Angle Bisector Theorem,

`(AE)/(EB) = (AD)/(DB)`     ...(1)

⇒ Also, in △ADC, DF bisects ∠ADC,

So, by the Angle Bisector Theorem,

`(AF)/(FC) = (AD)/(DC)`     ...(2)

But BD = DC. Hence, from (1) and (2),

`(AE)/(EB) = (AF)/(FC)`

Therefore, the points E and F divide the sides AB and AC in the same ratio.

So, by the converse of the Basic Proportionality Theorem,

EF || BC

Hence proved.