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Multiple converses to a theorem. Let us see how multiple statements can be considered converses to the Midpoint Theorem and how Theorem 7 is one of them. To formulate a converse we should express the original statement in “If … then …” form. For a complex statement, there may be multiple ways to do that. The Midpoint Theorem starts with △ABC and points P and Q on sides AB and AC respectively. The theorem has two assumptions and two conclusions, which we have named for further discussion. Assumptions: (P MID) P is the midpoint of AB and (Q MID) Q is the midpoint of AC. Conclusions: (PRLL) PQ ∥ BC and (HALF) PQ = BC/2. Let us use these four named conditions to discuss various possible statements. (i) A natural converse of the Midpoint Theorem would be: “If (PRLL) and (HALF) are true, then (P MID) and (Q MID) are true.” Write out this statement fully. Experiment and see that it seems to be true! Prove the statement. (Hint: Try to run a proof of the Midpoint Theorem backwards.) (ii) To see Theorem 7 as a converse of the Midpoint Theorem, we first write the Midpoint Theorem as follows: “Suppose P is the midpoint of side AB of △ABC and Q is a point on side AC. If Q is the midpoint of AC then PQ ∥ BC.” (Also PQ = BC/2, but let us set that aside for now.) Write the converse of the second sentence in quotes by keeping the first sentence the same. Verify that Theorem 7 is what you get! (And PQ = BC/2 also follows.) Now write Theorem 7 in terms of the four named conditions. (iii) The discussion so far suggests that it is reasonable to take two of the four listed statements as assumptions and ask if the other two are true. Verify that only one such combination remains to be examined. “If (P MID) and (HALF) are true, then can we conclude (Q MID) and/or (PRLL)?” Write this out in words. This is a precise version of question (1) stated before Theorem 7. Can you answer it?
(i) "If a segment PQ with P on AB and Q on AC is parallel to BC and has half the length of BC, then P and Q are the midpoints of AB and AC." Draw CR ∥ AB meeting PQ extended to form a parallelogram, proving congruence and midpoints. (ii) "Suppose P is the midpoint of side AB of △ABC and Q is a pointRead more
(i) “If a segment PQ with P on AB and Q on AC is parallel to BC and has half the length of BC, then P and Q are the midpoints of AB and AC.” Draw CR ∥ AB meeting PQ extended to form a parallelogram, proving congruence and midpoints.
(ii) “Suppose P is the midpoint of side AB of △ABC and Q is a point on side AC. If PQ ∥ BC, then Q is the midpoint of AC.” In named conditions: If (P MID) and (PRLL), then (Q MID) and (HALF).
(iii) “If P is the midpoint of AB and PQ = BC/2, must Q be the midpoint of AC and PQ ∥ BC?” No; swinging segment PQ around P can intersect AC at another location where it is neither parallel to BC nor bisecting AC.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-12/
See lessValidity of tiling methods. Show using reasoning that each of the three tiling methods we saw produces a tiling of the plane using the given 4-gon. You have to prove that when we place new copies using any of the three procedures, the entire plane is covered with no gaps and no overlaps. First, think carefully about what it means to prove this. Then find a proof for each procedure.
A valid tiling requires edge lengths to match identically and all interior angles meeting at every vertex to sum to 360° without gaps or overlaps. Method 1: Half-turn rotations around edge midpoints bring each of the four angles of the 4-gon together at every vertex, summing to 360° with matching adRead more
A valid tiling requires edge lengths to match identically and all interior angles meeting at every vertex to sum to 360° without gaps or overlaps.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-12/
See lessWhat fraction of the square is shaded?
The shaded region forms a square that is 1/5 of the total area. Each pair of parallel segments has slopes 1/2 and -2, making them perpendicular. By the Midpoint Theorem and congruent right triangles, the surrounding four corner regions can be disassembled and rearranged into four squares identical tRead more
The shaded region forms a square that is 1/5 of the total area. Each pair of parallel segments has slopes 1/2 and -2, making them perpendicular. By the Midpoint Theorem and congruent right triangles, the surrounding four corner regions can be disassembled and rearranged into four squares identical to the central shaded square. Thus, the entire square equals five identical squares, leaving 1/5 shaded.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 12 Quadrilaterals Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-12/
See lessThe sum of two integers is +5 and their difference is -21. Find the two numbers.
Let the two numbers be represented by x and y. From the given conditions, we form the pair of linear equations: x + y = 5 and x - y = -21. Adding these two equations eliminates y, resulting in 2x = -16, which gives x = -8. Substituting x = -8 into x + y = 5 gives -8 + y = 5, leading to y = 13. Thus,Read more
Let the two numbers be represented by x and y. From the given conditions, we form the pair of linear equations: x + y = 5 and x – y = -21. Adding these two equations eliminates y, resulting in 2x = -16, which gives x = -8. Substituting x = -8 into x + y = 5 gives -8 + y = 5, leading to y = 13. Thus, the two numbers are -8 and 13.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-13/
See lessThe difference between two numbers is 26 and one number is three times the other. Find the numbers.
Let the larger number be x and the smaller number be y. According to the problem statement, the system of linear equations is x - y = 26 and x = 3y. Using the substitution method, replace x with 3y in the first equation: 3y - y = 26, which simplifies to 2y = 26, yielding y = 13. Consequently, x = 3(Read more
Let the larger number be x and the smaller number be y. According to the problem statement, the system of linear equations is x – y = 26 and x = 3y. Using the substitution method, replace x with 3y in the first equation: 3y – y = 26, which simplifies to 2y = 26, yielding y = 13. Consequently, x = 3(13) = 39. The required numbers are 39 and 13.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-13/
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