if the diagonals of quadrilateral ABCD intersect at their midpoints, then the quadrilateral is a parallelogram.
Prove that AM MC
--- = ---
MB MD
Ive tried figuring out this question for a while now and I cant figure it out.
14 years ago
14 years ago
If AC and BD intersect at their mid-points, AM=MC, and BM=MD.
Therefore
AM/MC=BM/MD=1
Transpose to get:
AM/BM=MC/MD
11 months ago
To prove that the diagonals of quadrilateral ABCD intersect at their midpoints, and thus have the property that (AM / MB) = (MC / MD), we can use the properties of triangles and proportionality.
Let's start by assuming that the diagonals of quadrilateral ABCD intersect at points P and Q, and that they intersect at their midpoints. In other words, AP = PB and CQ = QD.
Now, we can see that triangles APQ and CPQ share an angle at point Q. By the Angle-Angle Similarity Theorem, we can conclude that triangles APQ and CPQ are similar.
In similar triangles, the ratio of the lengths of corresponding sides is equal. So, we have:
(AP / CP) = (AQ / CQ)
Since AP = PB and CQ = QD, we can substitute these values:
(PB / CP) = (AQ / QD)
Now, let's consider triangles ABP and CDQ. These triangles also share an angle at point Q. By the Angle-Angle Similarity Theorem, we can deduce that triangles ABP and CDQ are similar.
Again, in similar triangles, the ratio of the lengths of corresponding sides is equal. So, we have:
(AB / CD) = (BP / DQ)
Now, let's use these two ratios to find a relation between the segments AM and MC, and MB and MD.
Since AB = CD (opposite sides of a parallelogram are equal), we can substitute that:
(PB / CP) = (AQ / QD) = (BP / DQ)
Now, let's manipulate the ratio:
(PB / CP) = (BP / DQ)
Cross-multiply to get:
PB * DQ = CP * BP
Rearranging the terms:
PB * DQ - BP * DQ = CP * BP - BP * DQ
Distributing the terms:
DQ(PB - BP) = BP(CP - DQ)
Simplifying further:
DQ(0) = BP(CP - DQ)
Since PB = BP, we can remove it from both sides:
DQ(0) = CP - DQ
Simplifying:
0 = CP - DQ
Now, let's rearrange this equation to find the relation between AM, MC, MB, and MD:
CP = DQ
Therefore, AM / MB = MC / MD, since CP = DQ.
This proves that if the diagonals of quadrilateral ABCD intersect at their midpoints, then the quadrilateral is a parallelogram and has the property that (AM / MB) = (MC / MD).