In the adjoining figure, ABCD is a || gm in which E and F are the midpoints of AB and CD respectively.
Question:
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.
Solution:
In parallelogram ABCD, we have:
AD || BC and AB || DC
AD = BC and AB = DC
AB = AE + BE and DC = DF + FC
AB = AE + BE and DC = DF + FC
∴ AE = BE = DF = FC
Now, DF = AE and DF || AE.
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)
Comments
Bla bla 🖤 Darling 😘💟
Feb. 23, 2025, 6:35 a.m.
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