The diagram below shows the last frame from the video, with some lines and labels added.

AB is perpendicular to AO. AB, BC and CD are all equal in length.

Consider triangles OCB and DCO. Since OC is perpendicular to BD (as this is part of the*carpenter's square*), the two triangles are right-angled. They share a side (OC), and so since BC = CD, they are congruent. Hence angle BOC is equal to angle COD.

Now consider triangle OAB. By construction, AB is perpendicular to AO, so this triangle is also right-angled. It shares a hypotenuse with triangle BOC, and since AB = BC, OAB and OCB are congruent too. So angle AOB is equal to angles BOC and COD, and the angle has been trisected.

AB is perpendicular to AO. AB, BC and CD are all equal in length.

Consider triangles OCB and DCO. Since OC is perpendicular to BD (as this is part of the

Now consider triangle OAB. By construction, AB is perpendicular to AO, so this triangle is also right-angled. It shares a hypotenuse with triangle BOC, and since AB = BC, OAB and OCB are congruent too. So angle AOB is equal to angles BOC and COD, and the angle has been trisected.