# SAS Criterion in Triangles

We have an SAS axiom for congruence of two triangles. Do we have a similar criterion for similarity? Reflect for a moment on why there should be such a criterion.

**SAS Criterion:** If two sides of one triangle are respectively proportional to two corresponding sides of another, and if the *included* angles are equal, then the two triangles are similar.

Note the emphasis on the word *included*. If the equal angle is a non-included angle, then the two triangles may not be similar. Consider the following figure:

It is given that

\[\begin{align}& \frac{{AB}}{{DE}} = \frac{{AC}}{{DF}} \end{align}\]

\[\angle A = \angle D \]

The SAS criterion tells us that \(\Delta ABC\) ~ \(\Delta DEF\). Let us see the justification of this.

**Proof:** Assume that AB > DE. Take a point X on AB such that AX = DE. Through X, draw segment XY parallel to BC, intersecting AC at Y:

Note that \(\Delta AXY\) ~ \(\Delta ABC\), and thus:

\[\frac{{AX}}{{AB}} = \frac{{AY}}{{AC}}\]

Now, we will show that \(\Delta AXY\) and \(\Delta DEF\) are congruent. It is given that:

\[\frac{{DE}}{{AB}} = \frac{{DF}}{{AC}}\]

Since AX = DE, we have:

\[\frac{{DE}}{{AB}} = \frac{{AX}}{{AB}} = \frac{{AY}}{{AC}} = \frac{{DF}}{{AC}}\]

Thus, AY = DF. By the SAS criterion, \(\Delta AXY\) ≡ \(\Delta DEF\). And since \(\Delta AXY\) ~ \(\Delta ABC\), this means that \(\Delta DEF\) and \(\Delta ABC\) are similar.

**Example 1:** Consider the following figure:

Suppose that OA × OB = OC × OD. Show that \(\angle A\) = \(\angle C\) and \(\angle B\) = \(\angle D\).

**Solution:** Using the information given to us, we have:

\[\frac{{OA}}{{OC}} = \frac{{OD}}{{OB}}\]

Thus, two pairs of sides of \(\Delta AOD\) and \(\Delta COB\) are respectively proportional. Also, \(\angle AOD\) = \(\angle COB\), because these are vertically opposite angles. This means that \(\Delta AOD\) ~ \(\Delta COB\), and so:

\(\angle A\) = \(\angle C\) and \(\angle B\) = \(\angle D\)

We can also prove the converse of this. Suppose that it is given to us that \(\angle A\) = \(\angle C\) and \(\angle B\) = \(\angle D\). Then, \(\Delta AOD\) and \(\Delta COB\) are equi-angular, and hence similar. This means that

\[\begin{align}& \frac{{OA}}{{OC}} = \frac{{OD}}{{OB}}\\& \Rightarrow OA \times OB = OC \times OD \end{align}\]

**Example 2:** Consider the following figure:

Find the value of \(\angle E\).

**Solution:** We note that the three sides of the two triangles are respectively proportional:

\[\begin{align}& \left\{ \begin{gathered}\frac{{DE}}{{AB}} = \frac{{4.2}}{6} = 0.7\\ \frac{{DF}}{{AC}} = \frac{{2.8}}{4} = 0.7\\ \frac{{EF}}{{BC}} = \frac{{3.5}}{5} = 0.7 \end{gathered} \right.\\&\quad\frac{{DE}}{{AB}} = \frac{{DF}}{{AC}} = \frac{{EF}}{{BC}} \end{align}\]

Thus, the two triangles are similar, which means that they are equi-angular. Note carefully that the equal angles will be:

\[\begin{array}{l} \angle A = \angle D = 55.77^0 \\ \angle C = \angle F = 82.82^0 \\ \angle B = \angle E \end{array}\]

Finally,

\[\begin{array}{l} \angle E = \angle B = 180^0 - (55.77^0 + 82.82^0 )\\ \qquad\qquad\quad= 41.41^0 \end{array}\]

**Example 3:** Consider two similar triangles, \(\Delta ABC\) and \(\Delta DEF\):

AP and DQ are medians in the two triangles respectively. Show that

\[\frac{{AP}}{{BC}} = \frac{{DQ}}{{EF}}\]

**Solution:** Since the two triangles are similar, they are equi-angular. This means that \(\angle B\) = \(\angle E\). Also,

\[\begin{align}& \frac{{AB}}{{DE}} = \frac{{BC}}{{EF}}\\& \Rightarrow \quad\frac{{AB}}{{DE}} = \frac{{BC/2}}{{EF/2}} = \frac{{BP}}{{EQ}} \end{align}\]

Hence, by the SAS criterion, \(\Delta ABP\) ~ \(\Delta DEQ\). Thus, the sides of these two triangles will be respectively proportional, and so:

\[\begin{align}& \frac{{AB}}{{DE}} = \frac{{AP}}{{DQ}}\\& \Rightarrow \quad\frac{{AP}}{{DQ}} = \frac{{BC}}{{EF}}\\& \Rightarrow \quad\frac{{AP}}{{BC}} = \frac{{DQ}}{{EF}} \end{align}\]