Question

ABCD is a trapezium in which AD || BC. M is the mid-point of CD. Through M, PQ is drawn || to AB with P on AD produced and Q on BC.

1. Show that ΔDPM ≅ ΔCQM
2. Prove that area of ΔABM = `1/2` × area of trapezium ABCD.

Answer 1

We are given:

• Trapezium ABCD with AD || BC
• M is midpoint of CD
• PQ || AB with P on AD extended and Q on BC

We are asked to:

1. Show ΔDPM ≅ ΔCQM
2. Prove Area(ΔABM) = `1/2` × Area(trapezium ABCD)

Show ΔDPM ≅ ΔCQM

Step 1: Observe triangles

• Consider ΔDPM and ΔCQM.
• PQ || AB, so by corresponding angles ∠DPM = ∠CQM
• M is midpoint of CD, so DM = MC
• PM || QC, so ΔDPM and ΔCQM are between parallel lines with equal lengths on opposite sides

Step 2: Use SAS congruence

In ΔDPM and ΔCQM:

1. DM = MC   ...(Given, M midpoint of CD)
2. ∠DPM = ∠CQM   ...(Alternate interior angles, PQ || AB)
3. PM = QC   ...(Segments between parallel lines are equal, by midpoint construction)

By SAS rule,

ΔDPM ≅ ΔCQM

Prove Area(ΔABM) = `bb(1/2)` × Area(trapezium ABCD)

Step 1: Draw median from midpoint

• M is midpoint of CD
• Draw line AM
• Trapezium ABCD area formula:
  Area(trapezium) = `1/2 xx (AB + DC) xx "height"`

Step 2: Divide trapezium into triangles

• Draw diagonal AC (or median AM)
• ΔABM + ΔAMC = ΔABC
• By congruence from step i, area(ΔDPM) = area(ΔCQM)
• Using triangle subtraction and parallel lines,
  Area(ΔABM) = `1/2` Area(ABCD)