Given:QT bisects PS and PQ is parallel to ST Prove:PQ is congruent to ST

Question

Given:QT bisects PS and PQ is parallel to ST

Prove:PQ is congruent to ST

For the prove I have to have the ctptc for last answer.

Answer 1

Here's a diagram to help you understand better:

-------------------------------

| P T |

| |\ /| |

| | \ / | |

| | \ / | |

| | R | |

| | / \ | |

| | / \ | |

| |/ \| |

| Q S |

--------------------------------

Let R be the point of intersection of QT and PS.

Since QT bisects PS as given, then PR = RS.

Since every line has a total angle sum of 180°, then:

m/_SRT = 180° - m/_PRT = 180° - (180° - m/_PRQ) = m/_PRQ.

However, write in your proof: m/_SRT = m/_PRQ by vertical angles.

Since PQ || ST as given, then m/_QPS = m/_PST,

also since they are alternate interior angles.

Therefore, by ASA (angle side angle), /_\PQR = /_\RST.

Finally, PQ = ST, due to CPCTC

(corresponding parts of congruent triangles are congruent).

Answer 2

Draw and label a diagram!!!

M is the point of intersection∠

∠SPQ ≅ ∠QPS Alternate interiors

∠QMP ≅ ∠SMT Vertical angles

PM ≅ ≅SM Bisected

ΔQMP ≅ ΔTMS ASA

PQ ≅ ST CPCTC

QED