Question

The diagonals of a parallelogram ABCD intersect at O. A line through O intersects AB at X and DC at Y. Prove that $OX=OY$.

Answer 1

Hint: We will first of all construct a neat and clean diagram with proper labels to proceed in our question. Now, to prove $OX=OY$, we would need to prove the congruence of the two triangles containing OX and OY. So, the first step should be to prove the congruence of triangles OAX and OCY. We shall proceed in this manner to get the validation of our proof.

Complete step by step answer:
It has been given that the ends of the parallelogram are A, B, C and D respectively. Also, a line through O intersects AB at X and DC at Y. Now, we have enough information required for the construction of our parallelogram. This parallelogram can be constructed as follows:

Now, we can see in the above figure that, to prove $OX=OY$, we need to prove the congruence of $△OAX\text{ and }△OCY$. This can be done as follows:
In the triangles OAX and OCY, we have:
$\Rightarrow \mathrm{\angle }AOX=\mathrm{\angle }COY$
This is true by the property of opposite angles subtended between two-line pairs.
Now, we have:
$\Rightarrow OA=OC$
This is true because diagonals of a parallelogram bisect each other.
And, at last we have:
$\Rightarrow \mathrm{\angle }OAX=\mathrm{\angle }OCY$
This is true by the property of alternate interior angles of two parallel lines.
Therefore, using these three equations, we can apply the angle-side-angle (ASA) postulate of congruence. Thus, we can write:
$\Rightarrow △OAX\cong △OCY$
Therefore, now we can say that:
$\Rightarrow OX=OY$
Hence, $OX=OY$ has been proved for parallelogram ABCD.

Note: Proper construction and labelling of diagrams is very important when going for a proof in mathematics. It helps us see things and data more clearly. So, the diagram should always be neat. Also, one should be very careful in using the correct postulate while proving congruence of two triangles.