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
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.
CRM
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