Assemble the proof by dragging tiles to the statements and reasons columns.

Answer:
Step-by-step explanation:
Given;
Δ QPR ≅ ∠ STR
i) [tex]\frac{QP}{ST} = \frac{QR}{SR} = \frac{PR}{TR}[/tex]
ii )∠ PRQ = ∠ TRS ( Both are 90° since QR ⊥ PT )
iii) ∠ RPQ = ∠ STR ( same side ratio from (i) )
Hence it is proved that
Δ PQR [tex]\sim[/tex] Δ TSR
The [tex]\triangle PQR[/tex] is similar to [tex]\triangle TSR[/tex] using the AAA properties for similar triangle.
Given to us:
[tex]\overline{QR} \perp \overline{PT}[/tex] , So, we can write that [tex]\angle PRQ = \angle TRS=90^o[/tex];
[tex]\angle QPR \cong \angle STR[/tex];
Also we know that sum of all the angles of a triangle is 180°. So,
[tex](\angle QPR +\angle PRQ +\angle RQP) = (\angle STR +\angle TRS + \angle RST) = 180^o\\[/tex]
[tex]\angle RQP = \angle RST[/tex]
Therefore, using the AAA (Angle Angle Angle) properties for similar triangle. All three pairs of corresponding angles are the same.
Hence, [tex]\triangle PQR \sim \triangle TSR[/tex].
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