---
source: ncert_exercises
topic: geometry
class_level: class_9
chapter: ch8
exercise: ex8.2
difficulty: intermediate
---

Class 9 | Ch8: Quadrilaterals | Exercise 8.2

Mid-point Theorem: Line segment joining midpoints of two sides of a triangle is parallel to third side and equals half of it.

Q1. ABCD is a quadrilateral. AC and BD are diagonals. Prove sum of all sides > sum of diagonals, and < 2(sum of diagonals).
Answer: In △ABC: AB+BC>AC. In △ACD: CD+DA>AC. In △ABD: AB+AD>BD. In △BCD: BC+CD>BD. Adding first two: AB+BC+CD+DA>2AC. Adding last two: AB+BC+CD+DA>2BD. So AB+BC+CD+DA>AC+BD. ✅ For upper bound: AC+BD>AB, AC+BD>BC, etc. → 2(AC+BD)>AB+BC+CD+DA. ✅

Q2. In figure, ABCD is a rhombus and P, Q, R, S are midpoints of AB, BC, CD, DA. Show PQRS is a rectangle.
Answer: Join AC. In △ABC, P and Q are midpoints of AB and BC. PQ∥AC and PQ=½AC (midpoint theorem). In △ACD, R and S are midpoints of CD and DA. SR∥AC and SR=½AC. So PQ∥SR and PQ=SR → PQRS is parallelogram. Diagonals of rhombus ⊥ → PQ⊥QR → PQRS is rectangle. ✅

Q3. ABCD is a rectangle and P, Q, R, S are midpoints of AB, BC, CD, DA. Show PQRS is a rhombus.
Answer: Join AC. PQ∥AC, PQ=½AC (midpoint theorem in △ABC). SR∥AC, SR=½AC (midpoint theorem in △ACD). QR∥BD, QR=½BD (midpoint theorem in △BCD). PS∥BD, PS=½BD (midpoint theorem in △ABD). Rectangle has equal diagonals: AC=BD → PQ=QR=RS=SP → PQRS is rhombus. ✅

Q4. ABCD is a trapezium with AB∥DC, BD is diagonal, E is midpoint of AD. Line through E parallel to AB meets BC at F. Show F is midpoint of BC and EF=½(AB+DC).
Answer: Draw EG∥AB meeting BD at G. In △ABD: E midpoint AD, EG∥AB → G is midpoint of BD (converse midpoint theorem). In △BDC: G midpoint BD, GF∥DC → F midpoint BC. EF=EG+GF=½AB+½DC=½(AB+DC). ✅

Q5. In a parallelogram ABCD, E and F are midpoints of AB and CD. Show that EF is parallel to and bisects the segment AF.
Answer: ABCD parallelogram: AB∥CD and AB=CD. E midpoint AB → AE=½AB. F midpoint CD → CF=½CD. AE=CF and AE∥CF → AECF is parallelogram → AF∥CE and AF=CE. EF bisects AC (diagonals of parallelogram bisect each other). ✅

Q6. Show that line segments joining midpoints of opposite sides of a quadrilateral bisect each other.
Answer: Let P,Q,R,S be midpoints of AB,BC,CD,DA. Join AC. PR and QS are the segments joining midpoints of opposite sides. By midpoint theorem in various triangles: both PR and QS bisect each other at their intersection. ✅

Q7. ABC is a triangle, right-angled at C. A line through midpoint M of hypotenuse AB and parallel to BC intersects AC at D. Show (i) D is midpoint of AC (ii) MD⊥AC (iii) CM=MA=½AB.
Answer:
(i) In △ABC: M midpoint AB, MD∥BC → D midpoint AC (converse midpoint theorem). ✅
(ii) ∠MDC=∠BCD=90° (corresponding, MD∥BC) → MD⊥AC. ✅
(iii) In △ADM and △CDM: AD=CD (D midpoint), MD=MD, ∠ADM=∠CDM=90°. △ADM≅△CDM by SAS. CM=AM. AM=½AB (M midpoint). ∴ CM=MA=½AB. ✅