in trapezium ABCD ,E and F be the midpoint of the diagonals AC and BD respectively then find EF
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Step-by-step explanation:
ABCD is a trapezium and E,F are mid-points of diagonal AC and BD
AB∥CD [ one par of opposite side is parallel in trapezium ]
In △CDF and △GBF
⇒ DF=BF [ Since, F is mid-point of diagonal BD ]
⇒ ∠DCF=∠BGF [ DC∥GB and CG is a transversal ]
⇒ ∠CDF=∠GBF [ DC∥GB and BD is a transversal ]
∴ △CDF≅△GBF [ By ASA congruence rule ]
⇒ CD=GB [ C.P.C.T ] ---- ( 1 )
In △CAG, the points E and F are the mid-points of AC and CG respectively.
∴ EF=
2
1
(AG)
⇒ EF=
2
1
(AB−GB)
From ( 1 )
⇒ EF=
2
1
(AB−CD)
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