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Q15. In the given figure, ABCD is a parallelogram in which AC is a diagonal. G, E and F are the mid-points AB, BC and AC respectively. If $\u2206$GEF is an equilateral triangle, then prove that parallelogram ABCD is a rhombus not square.

Considering $\u25b3ABC$,

as per the mid point theorem of the triangle, the line segment connecting the midpoints of two sides of the triangle is parallel to the third side and is congruent to one half of the third side.

Since, GF is connected to the mid points of AB and AC, thus as per mid point theorem,

$GF=\frac{1}{2}BC$

Similarly, as GE is connected to the mid points of AB and BC, thus as per mid point theorem,

$GE=\frac{1}{2}AC$

Similarly, as FE is connected to the mid points of AC and BC, thus as per mid point theorem,

$FE=\frac{1}{2}AB$

Now, as, $\u25b3GEF$ is an equilateral triangle in which all the sides are equal and all the angles are equal.

Thus,

$GE=GF=FE\phantom{\rule{0ex}{0ex}}whichimplies;\phantom{\rule{0ex}{0ex}}\frac{1}{2}AC=\frac{1}{2}BC=\frac{1}{2}AB\phantom{\rule{0ex}{0ex}}or,AC=BC=AB$

As, AC=BC=AB, hence, $\u25b3ABC$ is an equilateral triangle, therefore,

$\angle ABC=60\xb0=\angle ACB=\angle BAC$

and as ABCD is a parallelogram in which opposite sides and angles are always equal,

Thus,

$AB=DCandBC=AD\phantom{\rule{0ex}{0ex}}or,AB=BC=CD=AD\phantom{\rule{0ex}{0ex}}and\angle ABC=\angle ADC=60\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}assumofadjacentanglesinparalle\mathrm{log}ramis180\xb0\phantom{\rule{0ex}{0ex}}thus,\angle ABC+\angle BCD=180\xb0\phantom{\rule{0ex}{0ex}}or,\angle BCD=180\xb0-60\xb0=120\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Hence,\angle BCD=\angle BAD=120\xb0\left\{asoppositeanglesareequalinparalle\mathrm{log}ram\right\}$

Thus, ABCD is a parallelogram in which all the sides are equal but the angles are not 90.

Hence, ABCD is a rhombus not a square.

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