Congruency: Congruent Triangles

131. Parallel Sides → Congruent Triangles · AB ∥ CD and AB = CD ⇒ △AOB ≅ △DOC by ASA

Given AB ∥ CD and AB = CD, the crossing lines AD and BC meet at O. By the ASA congruency rule, △AOB ≅ △DOC — and so AO = OD and BO = OC.

When two segments are parallel and equal in length (AB ∥ CD with AB = CD), the crossing lines AD and BC always cut each other at the same point O — and the two opposite triangles △AOB and △DOC come out congruent by ASA. The payoff: AO = OD and BO = OC, so the diagonals bisect each other. This is the building block behind 'the diagonals of a parallelogram bisect each other.'

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Selina ICSE: Congruency: Congruent Triangles

What this lesson covers

Try to break it

Drag A or B to reshape the figure. AB stays equal and parallel to CD by construction, so the alternate angles ∠BAO = ∠CDO and ∠ABO = ∠DCO never break. With AB = CD as the side between those two angles, ASA always gives △AOB ≅ △DOC — and so AO = OD, BO = OC at every position. Try to drag the figure so the two triangles stop being congruent; impossible.

How you build it

Build two equal parallel segments joined by crossing diagonals.

Place point A — the top-left corner of the figure.
Place point B to the right of A. AB is the first of the two parallel segments.
Draw segment AB (Segment tool: click A, then B).
Place point C below segment AB — this gives the figure its height. C will be the bottom-left corner.
With the Parallel tool, click point C, then click on segment AB — this draws the guide line through C parallel to AB. D will sit on this line.
Place point D on the parallel line, on the same side as B (i.e., to the right of C, since B is to the right of A), at roughly the same distance from C as AB. Don't aim for pixel-perfect placement — close is good enough. The figure just needs AB ∥ CD (which the parallel guide handles) and AB ≈ CD (which you control by spacing D ≈ AB-length from C).
Draw segment CD (Segment tool: click C, then D). AB and CD are now the two equal, parallel sides.
Draw segment AD joining A (top-left) to D (bottom-right) — one of the two crossing diagonals.
Draw segment BC joining B (top-right) to C (bottom-left). AD and BC cross at O — and that's where the congruent triangles AOB and DOC meet.

The proof, step by step

Prove that O is the midpoint of both AD and BC — that is, AO = OD and BO = OC.

In △AOB and △DOC: AB = CD (Given).
∠BAO = ∠CDO (Alternate angles, since AB ∥ CD with AD as transversal).
∠ABO = ∠DCO (Alternate angles, since AB ∥ CD with BC as transversal).
∴ △AOB ≅ △DOC (by ASA Congruence Rule).
∴ AO = OD (CPCT).
∴ BO = OC (CPCT). Hence O is the midpoint of both AD and BC.

Worked example

In the given figure, AB // CD and AB = CD. If AB = 10 cm and ∠BAO = 40°, find the length of CD and the measure of ∠CDO.

From congruency ΔAOB ≅ ΔDOC, corresponding parts are equal. So CD = AB = 10 cm. Also, ∠CDO = ∠BAO = 40° (alternate angles / c.p.c.t.).

CD = 5 cm, ∠CDO = 40°
CD = 10 cm, ∠CDO = 40° — correct
CD = 10 cm, ∠CDO = 50°
CD = 10 cm, ∠CDO = 90°