Question

In fig. PQRS is a quadrilateral and T and U are respectively points on PS and RS such that
$PQ=RQ$
$\mathrm{\angle }PQT=\mathrm{\angle }RQU$ ………………………(i)
$\mathrm{\angle }TQS=\mathrm{\angle }UQS$ ………………………(ii)
Prove that $QT=QU$.

Answer 1

Hint: First of all, we will find the sum of angles of a quadrilateral and then using that we will find the triangles which are equivalent to each other. Then using the laws of triangles which is given as when two sides and one angle of two triangles are the same i.e. side angle side rule (SAS), then those two triangles are equivalent to each other. By using this, we will prove $QT=QU$.

Complete step-by-step answer:
Now, in question we are given that, $\mathrm{\angle }PQT=\mathrm{\angle }RQU$ and $\mathrm{\angle }TQS=\mathrm{\angle }UQS$, which can be written as,
$\mathrm{\angle }PQT=\mathrm{\angle }RQU$ ……………..(i)
$\mathrm{\angle }TQS=\mathrm{\angle }UQS$ ……………….(ii)

Now, on adding the terms of expression (i) and (ii) we will get,
$\begin{array}{rl}& \mathrm{\angle }PQT=\mathrm{\angle }RQU\\ & +\mathrm{\angle }TQS=\mathrm{\angle }UQS\end{array}$
$\Rightarrow \mathrm{\angle }PQT+\mathrm{\angle }TQS=\mathrm{\angle }RQU+\mathrm{\angle }UQS$ …………….(iii)
Now, from the figure it can be seen that ,
$\mathrm{\angle }PQS=\mathrm{\angle }PQT+\mathrm{\angle }TQS$ ………………..(iv)
And $\mathrm{\angle }SQR=\mathrm{\angle }RQU+\mathrm{\angle }UQS$ …………..(v)
Now, substituting the values of equation (iv) and (v) in expression (iii), we will get,
$\mathrm{\angle }PQS=\mathrm{\angle }SQR$ ………………..(vi)
Now, we can see that there are two triangles in the quadrilateral PQRS, which are,
$\mathrm{\Delta }PQS$ and $\mathrm{\Delta }RQS$
Now, in $\mathrm{\Delta }PQS$ and $\mathrm{\Delta }RQS$, two sides are common which can be given as,
$PQ=RQ$
And two angles are equal which we found in expression (iv) i.e. $\mathrm{\angle }PQS=\mathrm{\angle }SQR$
We are also given that $PQ=RQ$, using all these relations we can say that it satisfies the side angle side rule of triangle, which states that when two sides and one angle of two triangles are same then those two triangles are equivalent to each other, this can be written mathematically as,
$\mathrm{\Delta }PQS\cong \mathrm{\Delta }RQS$
Now, as two triangles of a quadrilateral are equivalent to each other, two opposite angles will also be equal, which can be given as,
$\mathrm{\angle }QPS=\mathrm{\angle }QRS$
Now, on observing the figure it can be seen that $\mathrm{\angle }P$ is a common angle in $\mathrm{\angle }QPS$ and $\mathrm{\angle }QPT$, in the same way $\mathrm{\angle }R$ is a common angle in $\mathrm{\angle }QRS$ and $\mathrm{\angle }QRU$. So, from this we can say that $\mathrm{\angle }QPT=\mathrm{\angle }QRU$ ……………………..(vii)
Now, in $\mathrm{\Delta }PQS$ and $\mathrm{\Delta }RQS$, we are given that,
Two sides are same i.e. $QP=QR$
And two angles are equal which we found in expression (ii) i.e. $\mathrm{\angle }PQT=\mathrm{\angle }QRU$ ,
We are also given that $\mathrm{\angle }PQT=\mathrm{\angle }RQU$
So, from all these relations we can say that it satisfies the angle side angle theorem, which states that in triangle when, two angles and one side are common then those two triangles are equivalent to them, which can be given mathematically as,
$\mathrm{\Delta }QPT\cong \mathrm{\Delta }QRS$
Now, as two triangles of a quadrilateral are equivalent to each other, two opposite sides will also be equal, which can be given as,
$QT=QU$
Hence, proved as given in the question.

Note: In such a type of question, students might make mistakes in considering the same angle and same sides of the triangle and they might get confused in considering the equivalency of the triangles and due to that the sum might get wrong results, students must be careful in selecting corresponding sides of the two triangles equivalent to each other..