Question

In a triangle ABC, P& Q are points on side AB and AC respectively such that PQ is parallel to BC. AD is median of BC. prove that AD is bisects PQ also.

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

CRM

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