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
       Â
                ...(i)
Now, In ∆AQE and ∆ACD
∠AQE = ∠ACD
[corresponding angles]
∠QAE = ∠CAD [common]
Therefore, by using A.A. condition
         Â
Comparing (i) and (ii), we have
        Â
But         Â
                    [∵  AD is a median]
         Â
Hence AD bisects PQ.
In the given Fig. DEFG is a square and ∠BAC = 90°. Prove that
(i) ∆AGF ~ ∆DBG.
(ii) ∆AGF ~ ∆EFC.
(iii) ∆DBG ~ ∆AEFC
(iv) DE2 = BD × EC.