Prove that in a parallelogram the line passing through the mid-point of a side and the point of intersection of its diagonals bisects the opposite side.
To prove that in a parallelogram, the line passing through the mid-point of a side and the point of intersection of its diagonals bisects the opposite side, we can use the concept of triangles and properties of parallelograms.
Let's consider a parallelogram ABCD, where AB || CD and AD || BC. Let E be the point of intersection of the diagonals AC and BD. Let F be the mid-point of side AD.
To prove that line EF bisects side BC, we need to show that EF divides BC into two equal segments.
Proof:
Step 1: Draw diagonal AC and diagonal BD.
Step 2: Since F is the mid-point of AD, we can say that AF = FD.
Step 3: Since ABCD is a parallelogram, we know that opposite sides are equal. Therefore, AB = CD.
Step 4: Now, consider triangles AFE and CFE.
Step 5: AF = FD (from step 2)
FE (common side)
AB = CD (from step 3)
Step 6: By using the Side-Side-Side (SSS) congruence criterion, we can conclude that triangle AFE is congruent to triangle CFE.
Step 7: If two triangles are congruent, then their corresponding sides are equal. Therefore, AE = CE.
Step 8: Since AE = CE, we can conclude that line EF bisects side BC.
Hence, we have proved that in a parallelogram, the line passing through the mid-point of a side and the point of intersection of its diagonals bisects the opposite side.
Answers & Comments
Answer:
To prove that in a parallelogram, the line passing through the mid-point of a side and the point of intersection of its diagonals bisects the opposite side, we can use the concept of triangles and properties of parallelograms.
Let's consider a parallelogram ABCD, where AB || CD and AD || BC. Let E be the point of intersection of the diagonals AC and BD. Let F be the mid-point of side AD.
To prove that line EF bisects side BC, we need to show that EF divides BC into two equal segments.
Proof:
Step 1: Draw diagonal AC and diagonal BD.
Step 2: Since F is the mid-point of AD, we can say that AF = FD.
Step 3: Since ABCD is a parallelogram, we know that opposite sides are equal. Therefore, AB = CD.
Step 4: Now, consider triangles AFE and CFE.
Step 5: AF = FD (from step 2)
FE (common side)
AB = CD (from step 3)
Step 6: By using the Side-Side-Side (SSS) congruence criterion, we can conclude that triangle AFE is congruent to triangle CFE.
Step 7: If two triangles are congruent, then their corresponding sides are equal. Therefore, AE = CE.
Step 8: Since AE = CE, we can conclude that line EF bisects side BC.
Hence, we have proved that in a parallelogram, the line passing through the mid-point of a side and the point of intersection of its diagonals bisects the opposite side.
Step-by-step explanation: