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# In the fig $AX$ and $CY$ are respectively the bisectors of the opposite angles of $A$ and $C$ of a parallelogram $ABCD$ . Show that $AX||CY$ .

Last updated date: 21st Jul 2024
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Answer
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Hint: As we know that a parallelogram is a type of quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and opposite angles of a parallelogram are of equal measure. There is another property that the diagonals of a parallelogram bisect each other and the diagonal divides the parallelogram into two congruent triangles.

Complete step-by-step answer:
Let us first draw the diagram according to the data given in the question:

Now by the property of a parallelogram we can say that $\angle A = \angle C$ ( Opposite angles of a parallelogram are equal).
We will now divide both the sides by $2$ i.e. $\dfrac{1}{2}\angle A = \dfrac{1}{2}\angle C$ , So it gives us that $\angle YAX = \angle YCX$ .
If the angles are equal then their halves are also equal.
Also we know that if two lines are parallel then the sum of their adjacent angles is equal to ${180^ \circ }$ .
So we can write $\angle AYC + \angle YCX = {180^ \circ }$ , because $YA||CX$ .
We have solved before that $\angle YAX = \angle YCX$ , by putting this in the equation we can write that: $\angle AYC + \angle YAX = 180$ .
Therefore we can say that $AX||CY$ because the interior angles on the same side of the transversal are supplementary.
Hence proved that $AX||CY$ .

Note: Before solving this kind of question we should have the clear concept of parallelograms and their properties. We should note that if any of the angles of a parallelogram is a right angle, then its other angles will also be a right angle. Area of the parallelogram is $base \times height$ . We know that there are three different types of parallelogram i.e. Square, Rectangle and Rhombus.