Question

In triangle ABC, P and Q are points on the sides AB and AC respectively such that PQ is parallel to BC. Prove that medium AD, drawn from A to BC, bisects PQ.

Answer 1

Given: ∆ABC in which P and Q are points on sides AB and AC respectively such that PQ || BC and AD is a median.

To Prove : AD bisects PQ.

Proof: In ∆APE and ∆ABD
∠APE = ∠ABD
[corresponding angles]
and    ∠PAE = ∠BAD [common]
Therefore, by using AA similar condition
                   ΔAPE~ΔABD 
=> AE/AD=PE/BD
Now, In ∆AQE and ∆ACD
∠AQE = ∠ACD
[corresponding angles]
∠QAE = ∠CAD [common]
Therefore, by using A.A. condition
                  ΔAQE~ΔACD
=>AE/AD=EQ/DC......(ii)
Comparing (i) and (ii), we have
         PE/BD = EQ/DC
but BD=DC
=>PE=EQ
 [∵  AD is a median]
                 
Hence AD bisects PQ.

Answer 2

Answer:

AD median  bisects  PQ proved .

Explanation:

Given , ABC is a triangle in which P and Q are points on he sides AB and AC .

And  AD  is a median to BC  .

Also PQ|| BC .

Let  us consider that AD median intersect PQ at G .

Step 1:

We have PQ || BC

⇒∠APG = ∠ B and ∠AQG = ∠ C

So from triangles ΔAPG  and ΔABD,

∠APG = ∠ABD

and ∠PAE = ∠BAD           (Common )

Therefore , we get Δ APG$\sim$ ΔABD

⇒$\frac{PG}{BD} = \frac{AG}{AD}$ ..........(i)

Similarly , Δ AQG $\sim$ΔACD  so from this we get ,

$\frac{QG}{CD} =\frac{AG}{CD}$ ........(ii)

Now from (i) and (ii) we get ,

$\frac{PG}{BD} =\frac{QG}{CD}$

⇒$\frac{PG}{BD} =\frac{QG}{BD}$     [where AD is the median  ,so BD= CD ]

PG = QG

Final answer :

Hence , here we proved that median AD bisects PQ .

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