Question

Show that the diagonals of a rectangle are equal.

Answer 1

Hint:
Here, we need to prove that the diagonals of a rectangle are equal. We will use the properties of a rectangle to prove that the two triangles (with one diagonal as a side) are congruent. Then, using the congruent parts of congruent triangles are equal, we can prove that the diagonals of the rectangle are equal.

Complete step by step solution:
First, we will draw a rectangle with its diagonals.

Here, $$ABCD$$ is the rectangle where $$AC$$ and $$BD$$ are the diagonals. $$M$$ is the intersection point of the diagonals $$AC$$ and $$BD$$.
We need to prove that $$AC=DB$$.
We will use the properties of a rectangle and congruence of triangles to prove that the diagonals of a rectangle are equal.
Now, we know that the opposite sides of a rectangle are always equal.
Therefore, we get
$$AB=CD$$ and $$BC=AD$$
We know that all the interior angles of a rectangle are right angles.
Therefore, we get
$$\mathrm{\angle }ABC=\mathrm{\angle }BCD=\mathrm{\angle }CDA=\mathrm{\angle }DAB={90}^{\circ }$$
Now, we will prove that the triangles $$ABC$$ and $$DCB$$ are congruent.
In triangles $$ABC$$ and $$DCB$$, we have
$$AB=CD$$ (Opposite sides of a rectangle)
$$\mathrm{\angle }ABC=\mathrm{\angle }DCB={90}^{\circ }$$ (Interior angles of a rectangle)
$$BC=CB$$ (Common side)
Therefore, by S.A.S. congruence criterion, the triangles $$ABC$$ and $$DCB$$ are congruent.
Now, the congruent parts of two congruent triangles are equal.
Therefore, since the triangles $$ABC$$ and $$DCB$$ are congruent, we get
$$AC=DB$$

Therefore, we have proved that the diagonals of the rectangle are equal.


Note:
We have used the S.A.S. congruence criterion to prove that the triangles $$ABC$$ and $$DCB$$ are congruent. According to the S.A.S. congruence criterion, if two corresponding sides of two triangles are equal, and the corresponding angle formed by those sides is also equal, then the two triangles are congruent.