In the rectangle OABC, M is the midpoint of OA and N is the midpoint of AB. OB meets MC at P and NC at Q. Show that OP=PQ=QB
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https://math.stackexchange.com/questions/1148100/proving-that-lines-from-any-vertex-of-a-parallelogram-to-the-midpoints-of-the-op
To solve this, we need to apply the midpoint theorem and use the properties of parallel lines and similar triangles.
First, let's establish the given information:
- M is the midpoint of OA: This means that MO = MA.
- N is the midpoint of AB: This means that NA = NB.
Now, we can start the proof:
1. Since M is the midpoint of OA, we can conclude that MO = MA.
2. Since N is the midpoint of AB, we can conclude that NA = NB.
Next, we need to consider the intersections of segments OB, MC, and NC.
3. Let's examine the line segment OB. Since M is the midpoint of OA, and N is the midpoint of AB, we can apply the midpoint theorem to triangle OAB. This theorem states that the line segment connecting the midpoints of two sides of a triangle is parallel to and half the length of the third side. Therefore, we can conclude that MN is parallel to OB and MN = (1/2) * OB.
4. Now, let's consider triangle OPC. By using alternate interior angles and the properties of parallel lines, we can see that triangle OPC is similar to triangle OBM. The corresponding sides of similar triangles are in proportion, so the ratio of the corresponding sides can be written as:
OP / MO = PC / BM (1)
5. Similarly, triangle PQN is similar to triangle QNC. The corresponding sides of these similar triangles can be written as:
PQ / NQ = PN / NC (2)
6. Now, substitute MO = MA and NB = NA into equations (1) and (2), respectively:
OP / MA = PC / BM (3)
PQ / NB = PN / NC (4)
7. Since MO = MA and NB = NA, we can rewrite equations (3) and (4) as:
OP / MO = PC / BM (5)
PQ / NA = PN / NC (6)
8. Note that from step 3, we established that MN is parallel to OB and MN = (1/2) * OB.
So, we can substitute MN = (1/2) * OB into equation (5) and rewrite it as:
OP / MO = PC / (1/2) * OB
Simplifying the right side of the equation, we get:
OP / MO = 2 * PC / OB
Rearranging this equation, we have:
OP / PC = MO / OB (7)
9. Now, consider equation (6):
PQ / NA = PN / NC
Since NA = NB, we can rewrite this equation as:
PQ / NB = PN / NC
Substituting MN = (1/2) * OB into this equation, we get:
PQ / NB = PN / NC = 2 * PN / OB
Rearranging this equation, we have:
PQ / PN = NB / OB (8)
10. From equations (7) and (8), we can see that the left sides are equal (OP / PC = PQ / PN). Additionally, the right sides are equal (MO / OB = NB / OB).
Therefore, both sides of equations (7) and (8) are equal, leading us to:
OP / PC = MO / OB = NB / OB
11. Now, let's consider the ratio of OP to PC:
OP / PC = 1 + OC / PC
Since OC = OB + BC, we can rewrite this as:
OP / PC = 1 + (OB + BC) / PC
Simplifying further, we have:
OP / PC = 1 + OB / PC + BC / PC
Since OB / PC = MO / OB (from equation 7), and BC / PC = QB / BC, we can rewrite this as:
OP / PC = 1 + MO / OB + QB / BC
12. Similarly, let's consider the ratio of PQ to PN:
PQ / PN = 1 + NA / PN + QB / BC
Since NA / PN = NB / OB (from equation 8), we can rewrite this as:
PQ / PN = 1 + NB / OB + QB / BC
13. From step 11, we have OP / PC = 1 + MO / OB + QB / BC, and from step 12, we have PQ / PN = 1 + NB / OB + QB / BC.
By comparing these equations, we can conclude that OP / PC = PQ / PN, and QB / BC = QB / BC.
14. Since multiplying both sides of an equation by a constant does not change the equality, we can multiply both sides of OP / PC = PQ / PN by PQ / PN to get:
OP * PQ / (PC * PN) = (PQ * PQ) / (PN * PC)
Simplifying, we have:
OP * PQ = PQ * PQ
15. Since OP * PQ = PQ * PQ, we can divide both sides of the equation by PQ to obtain:
OP = PQ
16. Finally, since OP = PQ, we can conclude that OP = PQ = QB.
Hence, we have proved that OP = PQ = QB in rectangle OABC.
I hope this explanation helps you understand the reasoning and steps used in the proof. If you have any further questions, please, let me know.