In a `triangleABC` P and Q are points on AB and AC respectively and PQ||BC. Prove that the median AD bisects PQ.

Explanation

GIVEN A `Delta ABC` in which P and Q are points on AB and AC respectively such that `PQ||BC and AD ` is the median, cutting `PQ` at E.
TO PROVE `PE=EQ`.

PROOF In `Delta APE and Delta ABD`, we have
`angle PAE= angle BAD` [ common]
`angle APE = angle ABD ` [ corrsponding `angle`]
`:. Delta APE ~ Delta ABD` [ buy AA- similarity]
But, in similar triangles, the corresponding sides are proportional.
`:. (AE)/(AD)=(PE)/( BD)`
In `Delta AEQ and Delta ADC`, we have
`angle QAE= angle CAD` [ common]
`angle AQE = angle ACD` [ corresponding angles]
`:. Delta AEQ ~ Delta ADC` [ by AA- similarity]
But, in similar triangles, the correponding sides are proportional.
`:. (AE)/(AD)=(EQ)/(DC) " "....(ii)`
From (i) and (ii), we get `(PE)/(BD)=(EQ)/(DC) " " [ "each equal to" (AE)/(AD)]`
But, `BD=DC " " [ :. AD "is the median"]`
`:. PE=EQ`