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 ADEF and BC at GP 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

∴ AE = BE = DF = FC   

NowDF = 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 DF 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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