Context: We see that the diagonals of a parallelogram need not be equal.

Q. Do they bisect each other (do they intersect at their midpoints)? Reason and/or experiment to figure this out.

Question

Context: We see that the diagonals of a parallelogram need not be equal. Q. Do they bisect each other (do they intersect at their midpoints)? Reason and/or experiment to figure this out.

Homi · Accepted Answer

The diagonals of a parallelogram always cut each other exactly in half.

Step 1 — Understanding Parallelogram Properties

A parallelogram is a four-sided shape. Its opposite sides are parallel to each other. This means side AB is parallel to side DC. Also, side AD is parallel to side BC. In a parallelogram, opposite sides are also equal in length. So, AB = DC and AD = BC. Opposite angles are also equal. So, $\angle A = \angle C$ and $\angle B = \angle D$ . From the given diagram, we see AB = 4 cm and DC = 4 cm. Also, AD = 5 cm and BC = 5 cm. We see $\angle A = 30^\circ$ and $\angle C = 30^\circ$ . We also see $\angle B = 150^\circ$ and $\angle D = 150^\circ$ . These measurements confirm that ABCD is a parallelogram.

Step 2 — Drawing Diagonals

Let us draw the two diagonals of the parallelogram. One diagonal connects point A to point C. The other diagonal connects point B to point D. Let these two diagonals cross each other at a point. We will call this intersection point O.

Diagram 1

Step 3 — Proving Triangle Congruence

We need to show that the diagonals cut each other in half. This means we need to show that AO = OC and BO = OD. Let us look at two triangles formed by the diagonals. Consider triangle AOB and triangle COD. Since side AB is parallel to side DC, and line AC is a transversal (a line crossing parallel lines). The alternate interior angles are equal. So, $\angle OAB = \angle OCD$ . (This is the same as $\angle CAB = \angle ACD$ ). Similarly, since side AB is parallel to side DC, and line BD is a transversal. The alternate interior angles are equal. So, $\angle OBA = \angle ODC$ . (This is the same as $\angle DBA = \angle BDC$ ). We already know that AB = CD because they are opposite sides of a parallelogram. Now, let's compare $\triangle AOB$ and $\triangle COD$ : We have: $\angle OAB = \angle OCD$ (Angle) AB = CD (Side) $\angle OBA = \angle ODC$ (Angle) By the ASA (Angle-Side-Angle) congruence rule, these two triangles are congruent.

$\boxed{\triangle AOB \cong \triangle COD}$

Step 4 — Concluding Bisection

Since $\triangle AOB$ is congruent to $\triangle COD$ , their corresponding parts are equal. This is known as CPCTC (Corresponding Parts of Congruent Triangles are Congruent). So, the side AO in $\triangle AOB$ is equal to the side CO in $\triangle COD$ . $AO = CO$ This means that point O is exactly the midpoint of the diagonal AC. Also, the side BO in $\triangle AOB$ is equal to the side DO in $\triangle COD$ . $BO = DO$ This means that point O is exactly the midpoint of the diagonal BD. Since point O is the midpoint of both diagonals, we can say that the diagonals bisect each other.

Answer

Yes. The diagonals of a parallelogram always bisect each other.

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