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A line through the point P on side XY of ΔXYZ is drawn parallel to YZ, meeting XZ at Q. Here XP = 4.8 cm and PY = 7.2 cm. A student objects that BPT cannot be used because XP ≠ PY. If XZ = 25 cm, find XQ.

Class 10Triangles (Similarity)Geometry — Similarity
  • A10 cm
  • B12.5 cm
  • C15 cm
  • D16.7 cm

Step-by-Step Solution

BPT applies to ANY line parallel to a side — the midpoint case is only the special instance where the ratio happens to be 1 : 1, so the objection is wrong. cm, and , so cm. 12.5 cm is the student’s midpoint assumption; 15 cm is , the other part; 16.7 cm applies to the whole of instead of splitting it into 2 + 3 parts.
Answer: A) 10 cm

Key idea: Three similarity criteria

**AA (Angle-Angle)**: Two angles equal → similar. (Third angle is automatic by 180° angle sum.)

**SSS (Side-Side-Side proportionality)**: All three pairs of corresponding sides in same ratio → similar.

**SAS (Side-Angle-Side)**: Two pairs of sides proportional AND included angles equal → similar.

**Important**: SSA does NOT prove similarity (just like for congruence).

AA is by far the most-used in problems: spot two equal angles (often via parallel lines or shared angle) and declare similarity.

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