Answer:
Given : [tex]EP\parallel TQ[/tex]
And, [tex]EQ \cap PT =K[/tex]
That is, K is the intersection point of the line segment EQ and PT.
To prove: [tex]\triangle TKQ\sim \triangle PKE[/tex]
Proof:
Here, [tex]EP\parallel TQ[/tex]
And, QE is the common transversal of these parallel lines,
Thus, By the alternative interior angle theorem,
[tex]\angle KEP\cong \angle KQT[/tex]
Similarly, [tex]\angle KPE\cong \angle KTQ[/tex]
Thus, By AA similarity postulate,
[tex]\triangle TKQ\sim \triangle PKE[/tex]
Hence, Proved.
