Question

In figure \[\dfrac{{QR}}{{QS}} = \dfrac{{QT}}{{PR}}\]and\[\angle 1 = \angle 2\]. Show that \[\vartriangle PQS \sim \vartriangle TQR\]

Answer 1

Hint: We prove two triangles similar by SAS theorem of similarity. Use the given set of equal angles to prove the triangle PQR an isosceles triangle. Use the proportional sides and common angle to show triangles similar.
SAS theorem of similarity states that if two sides in one triangle are proportional to two sides in another triangle and the included angle in both are equal, then the two triangles are similar.
Two triangles are equilateral if they have two sides equal. Also, angles opposite to equal sides are equal.

Complete step-by-step solution:
We are given \[\angle 1 = \angle 2\]
In \[\vartriangle PQR\], \[\angle 1 = \angle 2\]
Since we know the sides opposite to equal angles are equal in any triangle.
\[ \Rightarrow PQ = PR\]...............… (1)
So\[\vartriangle PQR\] is an isosceles triangle.
Now we are given \[\dfrac{{QR}}{{QS}} = \dfrac{{QT}}{{PR}}\]
Substitute the value of\[PR = PQ\] in the proportion
\[ \Rightarrow \dfrac{{QR}}{{QS}} = \dfrac{{QT}}{{PQ}}\].................… (2)
Now in \[\vartriangle PQS\] and \[\vartriangle TQR\];
 \[\dfrac{{QR}}{{QS}} = \dfrac{{QT}}{{PQ}}\]{From equation (2)}
\[\angle A = \angle A\](Common angle in both triangles)
We have two sides in one triangle that are proportional to two sides in another triangle and included angles equal in both the triangles.
Then by SAS similarity theorem, we can say that \[\vartriangle PQS \sim \vartriangle TQR\]
\[\therefore \vartriangle PQS \sim \vartriangle TQR\]

Note: Students many times confuse SAS similarity theorem with SAS congruence. They try to prove sides of triangles congruent and equal, but keep in mind here we have to prove triangles similar and the rules are different for proving congruence and different for proving similarity. Also, focus on the pair of equal angles in both triangles as they should be between the proportional sides, i.e. the angle should be included and not any of the angles.