Math, asked by SNI, 1 year ago

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.

Answers

Answered by sonabrainly
13

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.





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2

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