Answer: Yes ΔRWS≈ΔQWT.
Step-by-step explanation:
We know that when two chords intersect each other inside a circle, the products of their segments are equal.
Therefore, [tex]QW\times WS=TW\times RW\\\Rightarrow\frac{QW}{RW}=\frac{TW}{WS}[/tex]
Now in triangle QWT and triangle RWS
∠QWT=∠RWS [Vertically opposite angles]
[tex]\frac{QW}{RW}=\frac{TW}{WS}[/tex]
Therefore, By SAS similarity rule
ΔRWS≈ΔQWT
Therefore, yes ΔRWS≈ΔQWT.
