Respuesta :

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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