Question

ABCD is a parellelogram, E and F are the mid points AB and CD respectively. GH is any line intersecting AD, EF and BC at G, P and H respectively. prove that GP = PH.

Answer 1

In ||gm ABCD, E and F are mid -points of AB and CD

GH is any line intersecting AD, EF and BC at GP and H respectively

To prove : GP = PH

Proof : $\because$ E and F are the mid -points of AB and CD

$\therefore AE=EB=\frac{1}{2}AB$ and

CF = FD = $\frac{1}{2}CD$

But AB = CD (opposite sides of a ||gm)

$\therefore \frac{1}{2}AB=\frac{1}{2}CD$

$\Rightarrow EB=CF$

But EB|| CF \( (\because AB || CD)

$\therefore$ BEFC is a ||gm

$\therefore$ BC ||EF and BE = PH ....(i)

Now $\because$ BC || EF

$\therefore$ AD || EF

\(\because BC || AD, opposite sides of a ||gm)

$\therefore$ AEFD is a || gm

$\therefore$ AE = GP ....(ii)

But E is the mid-point of AB

$\therefore$ AB = BE ....(iii)

From (i), (ii) and (iii)
GP = PH