Question

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

Answer 1

Answer:

Solution:

Given ABCD is a parallelogram

The diagram is attached below

The diagonal of a parallelogram ABCD intersect at O

in parallelogram the diagonal bisects each other

The diagonal in the given figure are AC and BD which intersect at O

If we consider two triangles, ∆ AOX and ∆ YOC

Then we get following similarities:-

AO = OC (since, diagonals of parallelogram bisects each other)

Angle AOX = Angle COY ( Vertically opposite angles)

Angle XAO = Angle YCO (Alternate Interior angles)

By ASA Theorem (Angle-Side-Angle)

The Angle Side Angle Postulate (ASA) says triangles are congruent if any two angles and their included side are equal in the triangles.

\Delta A O X \cong \Delta Y O CΔAOX≅ΔYOC

So, By CPCT (Corresponding Part of Congruent Triangles)

“Corresponding Parts of Congruent Triangles” theorem states that if we take two or more triangles which are congruent to each other then the corresponding angles and the sides of the triangles are also congruent to each other i.e., their corresponding parts are equal to each other

We get, OX = OY

Hence proved