Question

please please give me answer

Answer 1

hi,
i) In Δ AEX & ΔBEC
         
            AE=BE (GIVEN)
            XE=CE (    ''      )
       ∠AEX=∠BEC (VERTICALLY OPPOSITE ANGLES)
BY SAS BOTH ARE CONGRUENT

ii) SIMMILARLY IN Δ AFY & ΔCFB 
    BY SAS BOTH ARE  CONGRUENT
II) IF WE JOIN EF BY MIDPOINT THEOREM IT WILL BE PARALLEL TO BC.

WHEN WE CONSIDER ALL THREE LINES EF,BC & XAY 
THERE ARE TWO TRANSVERSALS, AB,AC
INTERCEPTS ARE EQUAL ie AE=BE , AF=FC.
SO, BY INTERCEPT THEOREM ALL LINES ARE PARALLEL
HENCE PROOVED