Two parallel lines l and m are intersected by another pair of parallel lines p and q, forming a classic geometry setup often seen in textbooks like NCERT Class 9. Without the figure or options, this typically tests congruence of triangles ABC and CDA (where points A, B, C, D form at intersections), or angle relationships like alternate interior angles being equal.

Key Geometry Properties

Parallel lines cut by transversals create equal angles and congruent segments. Here's what's always true:

  • Corresponding angles equal : Angles in matching positions (e.g., top-left at each intersection).
  • Alternate interior angles equal : Angles on opposite sides of the transversal, between parallels (e.g., ∠4 = ∠5).
  • Consecutive interior angles supplementary : Add to 180° on the same side.

Since p || q adds another parallel pair, segments between l and m are equal along p and q.

Proving Triangle Congruence (Common Question)

This matches NCERT Ex 7.1 Q4: Prove ΔABC ≅ ΔCDA (A on p∩l, B on p∩m, C on q∩m, D on q∩l).

Step-by-Step Proof (ASA Criterion) :

  1. l || m, transversal AC → ∠ACB = ∠CAD (alternate interior angles).
  1. p || q, transversal AC → ∠BAC = ∠ACD (alternate interior angles).
  1. AC common side.
  2. Thus, ΔABC ≅ ΔCDA by ASA (two angles + included side).

Aspect| Why Equal/Congruent 110
---|---
∠BCA & ∠DCA| Alternate interior (l
∠BAC & ∠CAD| Alternate interior (p
Side AC| Common
Result| ΔABC ≅ ΔCDA (sides AB=CD, BC=DA)

Likely Correct Option

If options involve:

  • Triangle congruence: ΔABC ≅ ΔCDA is correct.
  • Angles: Alternate interior equal, or same-side supplementary.
  • Avoid: Vertical angles only (not key here), or non-parallel assumptions.

TL;DR : Correct option proves ΔABC ≅ ΔCDA via ASA, or states equal alternate angles—standard for this crisscross parallel setup.

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