State true or false.

In figure, G is the point of concurrency of medians of $△ DEF$ . Take point H on ray DG such that DG=GH Then the quadrilateral GEHF is a parallelogram.

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True

False

answer is A.

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Detailed Solution

Suppose A is the point at which the median DG cuts EF.
Question Image

As, median cuts the side in half, so, the point A is the midpoint of EF.
So,

AE = AF \dots \dots (1)

.
Also, median's concurrence point= G.
Now, every median intersects at single location when the drawn from each triangle vertex.
This common point divides median in a ratio of 2:1.
Thus

DG : GA = 2 : 1

\frac{DG}{GA} = \frac{2}{1}

DG = 2 AG

DG = 2 AG \dots \dots (2)

Since,

DG = GH

So,

GH = 2 AG

Hence, A =mid-point of GH
So,

GA = AH

Thus, A =mid-point of GH
Therefore,

GA = AH

As, diagonals bisect each other so quadrilateral must be parallelogram.
In quadrilateral GEHF diagonals GH and EF meet at point A.
Thus, A bisects both diagonals for the above discovered midpoints.
Therefore, GEHF = parallelogram.
So, the given statement is true.
Correct option is 1.

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