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🔷 Chapter 12: Quadrilaterals • Exercise Set 12.3

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Exercise Set 12.1 4Exercise Set 12.2 4Exercise Set 12.3 3Exercise Set 12.4 1End-of-Chapter Exercises 7All Exercises 19
Q1

(i) If P, Q, R are the midpoints of sides AB, AC, BC respectively of ΔABC, show that ΔPQR is congruent to ΔQPA and to two other triangles which you should identify.
(ii) Suppose someone erases ΔABC, leaving only ΔPQR on the paper. Can you reconstruct ΔABC from ΔPQR?

  • (i) By the Midpoint Theorem, PQ || BC, QR || AB, and RP || AC. Because PQ || BR and PR || BQ, PBQR is a parallelogram. The diagonal PR divides it into two congruent triangles: ΔPQR ≅ ΔRBP. By the same logic for the other parallelograms, ΔPQR ≅ ΔQPA ≅ ΔCQR.
  • (ii) Yes, you can reconstruct ΔABC. To find the original vertices A, B, and C, simply draw a line through point P parallel to QR, a line through Q parallel to PR, and a line through R parallel to PQ. The intersections of these three lines will perfectly recreate the original ΔABC.
A B C P Q R
Medial triangle ΔPQR divides ΔABC into 4 congruent triangles: ΔPQR ≅ ΔQPA ≅ ΔRBP ≅ ΔCQR
Q2

In ΔABC, let M and N be midpoints of AB and AC respectively. Let D be any point on BC. Show that MN bisects AD.

Proof:
1. By the Midpoint Theorem, MN connects the midpoints of AB and AC, therefore MN || BC.
2. Let the segment AD intersect MN at point X.
3. In ΔABD, M is the midpoint of AB, and the segment MX is part of the line MN, which is parallel to the base BD (since MN || BC).
4. By the Converse of the Midpoint Theorem (Theorem 7), a line drawn through the midpoint of one side of a triangle parallel to the base will bisect the third side.
5. Therefore, MX bisects AD, meaning X is the midpoint of AD. Thus, MN bisects AD.

A B C M N D X
Midsegment MN ∥ BC bisects segment AD at X (AX = XD)
Q4

Suppose the midpoints of sides AB, BC, CD and DA of a quadrilateral ABCD are P, Q, R and S respectively.
(i) Show that PR and QS bisect each other.
(ii) Show that if AC = BD, then PR and QS are perpendicular. Is the converse true?

  • (i) By Varignon's Theorem (Theorem 9), joining the midpoints of any quadrilateral forms a parallelogram. Thus, PQRS is a parallelogram. In any parallelogram, the diagonals bisect each other. Therefore, PR and QS bisect each other.
  • (ii) If the diagonals of the original quadrilateral are equal (AC = BD), then the sides of the Varignon parallelogram PQRS, which are half the lengths of the diagonals (PQ = SR = AC/2 and PS = QR = BD/2), must be equal to each other. A parallelogram with equal adjacent sides is a rhombus. The diagonals of a rhombus are perpendicular. Thus, PR ⊥ QS. Yes, the converse is true: If PR ⊥ QS, then PQRS is a rhombus, meaning its sides are equal, which forces AC = BD.
A B C D P Q R S
Varignon Parallelogram PQRS: Diagonals PR and QS bisect each other