### Video Transcript

Is polygon π΄π΅πΆπ· similar to
polygon πΊπΉπΈπ?

From the figure, we can see that
the two polygons weβve been given are each parallelograms. So we can conclude that they are at
least the same type of shape to begin with. To determine whether theyβre
similar, we need to test two things. Firstly, we need to test whether
corresponding pairs of angles are congruent. And secondly, we need to test
whether corresponding pairs of sides are in proportion or in the same ratio.

Now, itβs important to remember
that when weβre working with similar polygons, the order of the letters is
important. So if these polygons are similar,
then the angle at π΄ will correspond to the angle at πΊ. The angle at π΅ will correspond to
the angle at πΉ, and so on. We can therefore deduce that the
polygons have been drawn in the same orientation.

Letβs consider the angles first of
all then. In polygon πΊπΉπΈπ, weβve been
given a marked angle of 110 degrees. And in polygon π΄π΅πΆπ·, weβve been
given a marked angle of 70 degrees. One thing we do know about
parallelograms is that their opposite angles are equal. So in parallelogram πΊπΉπΈπ, the
angle at π will be 110 degrees. And in parallelogram π΄π΅πΆπ·, the
angle at πΆ will be 70 degrees.

Letβs consider the angle at πΊ in
the polygon πΊπΉπΈπ. By extending the line πΉπΊ, we now
see that we have two parallel lines ππΊ and πΈπΉ and a transversal πΊπΉ. We know that corresponding angles
in parallel lines are equal, which means that the angle above the line ππΊ will be
equal to the angle above the line πΈπΉ. Itβs 110 degrees. We also know that angles on a
straight line sum to 180 degrees, which means the angle below ππΊ will be 180 minus
110. Itβs 70 degrees. This shows us that the angle at πΊ
in polygon πΊπΉπΈπ is equal to the angle at π΄ in polygon π΄π΅πΆπ·.

We already said that opposite
angles in parallelograms are equal. So the angle at πΈ is also 70
degrees, which is equal to the angle at πΆ in polygon π΄π΅πΆπ·. We could use the same logic in
parallelogram π΄π΅πΆπ· to show that the angles at π΅ and π· are each 110 degrees,
which are the same as the angles at πΉ and π in the larger polygon. Weβve shown then that all pairs of
corresponding angles are indeed congruent. So our answer to the first check is
yes.

Letβs now consider whether
corresponding pairs of sides are in proportion. Firstly, from the figure, we can
see weβve been given side length π΄π΅; itβs 13 centimeters. And if the two polygons are
similar, this will correspond to πΊπΉ. We havenβt been given the length of
πΊπΉ. But we know that opposite sides in
a parallelogram are equal in length. So it will be the same as ππΈ. Comparing the ratio of these two
sides then, we find that π΄π΅ over πΊπΉ is equal to 13 over 26, which simplifies to
one-half.

The other pair of potentially
corresponding sides weβve been given are π΅πΆ and πΈπΉ. The ratio here is 11.5 over 23,
which again simplifies to one-half. Now, Iβve written πΈπΉ here. But really, if weβre to be
consistent with the order of letters, then we should really write πΉπΈ as point πΉ
corresponds to point π΅ and point πΈ corresponds to point πΆ. However, in calculating the ratio,
the length πΈπΉ is of course the same as the length πΉπΈ. So it makes no practical
difference.

The ratio of these pairs of
corresponding sides is therefore the same. As opposite sides in a
parallelogram are equal in length, the same will be true for the remaining two pairs
of corresponding sides. And so the answer to our second
check, βare corresponding pairs of sides in proportion?β, is also yes. Hence, both of the criteria for
these two polygons to be similar are fulfilled. And so we can answer, yes, polygon
π΄π΅πΆπ· is similar to polygon πΊπΉπΈπ.