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.

Answer 1

In parallelogram ABCD, we have:
AD || BC and AB || DC

AD = BC and AB = DC
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)