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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
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
maithilimundle:
hey are you fine wid d solution ??
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