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△rtp with xs is shown. given: xs ⊥ rt and rx ⊥ xt prove: △xtr ~ △sxr co…

Question

△rtp with xs is shown.
given: xs ⊥ rt and rx ⊥ xt
prove: △xtr ~ △sxr
complete the two - column proof with the provided statements and reasons.
∠xsr ≅ ∠txr aa ~
△xtr ~ △sxr definition of congruence
∠xrs ≅ ∠trx definition of perpendicular lines
m∠xsr = m∠txr reflexive property of congruence
m∠xsr = 90° and m∠txr = 90° transitive property of equality
∠xsr and ∠txr are right angles definition of right angles

Explanation:

Step 1: Definition of perpendicular lines

Since \(XS\perp RT\) and \(RX\perp XT\), by the definition of perpendicular lines, \(m\angle XSR = 90^{\circ}\) and \(m\angle TXR=90^{\circ}\)

Step 2: Definition of right angles

Angles with a measure of \(90^{\circ}\) are right angles. So, \(\angle XSR\) and \(\angle TXR\) are right angles

Step 3: Definition of congruence

If two angles have the same measure (\(m\angle XSR = m\angle TXR = 90^{\circ}\)), then \(\angle XSR\cong\angle TXR\) (by the definition of congruence for angles)

Step 4: Reflexive property of congruence

For any angle \(\angle R\), \(\angle R\cong\angle R\) (reflexive property of congruence)

Step 5: AA (Angle - Angle) similarity criterion

In \(\triangle XTR\) and \(\triangle SXR\), we have two pairs of congruent angles (\(\angle XSR\cong\angle TXR\) and \(\angle R\cong\angle R\)). By the AA similarity criterion, \(\triangle XTR\sim\triangle SXR\)

Answer:

  1. \(m\angle XSR = 90^{\circ}\) and \(m\angle TXR = 90^{\circ}\)
  2. \(\angle XSR\) and \(\angle TXR\) are right angles
  3. \(\angle XSR\cong\angle TXR\)
  4. \(\angle R\cong\angle R\)
  5. \(\triangle XTR\sim\triangle SXR\)
  6. (No additional statement needed as the proof is complete with the given steps for the similarity of \(\triangle XTR\) and \(\triangle SXR\))