Answer:
QPxMO is equal to QOMN
Explanation:
Since we are given that angle QPO is congruent to angle MNO, we can conclude that triangle QPO is similar to triangle MNO.
Therefore, we can write:
QP / MN = PO / NO (corresponding sides of similar triangles)
Multiplying both sides by MN, we get:
QP = (PO / NO) * MN
Now, we can use this expression for QP to prove that QPxMO=QOMN:
QPxMO = QP + PO (by segment addition postulate)
QPxMO = [(PO / NO) * MN] + PO (substituting QP with (PO / NO) * MN)
QPxMO = PO * [(MN + NO) / NO] (factoring out PO)
QPxMO = PO * (MO / NO) (since MN + NO = MO)
Now, we can see that PO * (MO / NO) is the same as QOMN, which means:
QPxMO = QOMN
Therefore, we have proven that QPxMO is equal to QOMN.
Hope this helped, I'm sorry if it is wrong. If you need more help, ask me!
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