In the adjoining figure, ABCD is a || gm in which E and F are the midpoints of AB and CD respectively.
In the adjoining figure, ABCD is a || gm in which E and F are the midpoints of AB and CD respectively. If GH is a line segment that cuts AD, EF and BC at G, P and H respectively, prove that GP = PH.
In parallelogram ABCD, we have:
AD || BC and AB || DC
AB = AE + BE and DC = DF + FC
∴ AE = BE = DF = FC
i.e., AEFD is a parallelogram.
∴ AD || EF
Similarly, BEFC is also a parallelogram.
∴ EF || BC
∴ AD || EF || BC
Thus, AD, EF and BC are three parallel lines cut by the transversal line DC at D, F and C, respectively such that DF = FC.
These lines AD, EF and BC are also cut by the transversal AB at A, E and B, respectively such that AE = BE.
Similarly, they are also cut by GH.
∴ GP = PH (By intercept theorem)
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