Similarity (With Applications to Maps and Models)

317. The SSS Similarity Rule · Proportional sides mean identical shapes

The ratio of corresponding sides remains constant, and corresponding angles are equal.

PQRABCDrag K to scale ΔABCPQ = 360.6PQ = 360.6QR = 400QR = 400PR = 360.6PR = 360.6AB = 180.3AB = 180.3BC = 200BC = 200AC = 180.3AC = 180.3∠P = 67°∠P = 67°∠Q = 56°∠Q = 56°∠R = 56°∠R = 56°∠A = 67°∠A = 67°∠B = 56°∠B = 56°∠C = 56°∠C = 56°k = AB/PQ = 0.5k = AB/PQ = 0.5 = AC/PR = 0.5 = AC/PR = 0.5 = BC/QR = 0.5 = BC/QR = 0.5K
SSS similarity: if the three pairs of corresponding sides are proportional (all in the same ratio), the triangles are similar, and their corresponding angles are automatically equal.

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Selina ICSE: Similarity (With Applications to Maps and Models)

What this lesson covers

Try to break it

Drag K to resize △ABC. △PQR scales in response, and the three side ratios AB/PQ, BC/QR, CA/RP always read the same number — that's the SSS similarity condition. Try to drag K so one ratio drifts away from the others; impossible.

How you build it

Build a triangle with all sides half of PQR, and see why SSS makes them similar.

  • Point tool: mark P near the top.
  • Point tool: mark Q below and to the right of P.
  • Point tool: mark R below and to the left, so PQR is a clear triangle.
  • Segment tool: join P to Q.
  • Segment tool: join Q to R.
  • Segment tool: join R to P. PQR is the reference triangle.
  • Midpoint tool: click P then Q — it drops M exactly halfway, so PM is exactly half of PQ. No ruler needed.
  • Midpoint tool: click P then R — N lands exactly halfway, so PN is exactly half of PR.
  • Segment tool: join M to N. By the midpoint theorem MN is exactly half of QR — so all three sides PM, PN, MN are half of PQ, PR, QR. Equal ratios on every side means △PMN ~ △PQR by SSS, and the angles automatically match.

The proof, step by step

Prove that two triangles with all sides in the same ratio are similar (SSS).

  • Given: AB/PQ = BC/QR = AC/PR = k.
  • Mark point D on AB such that AD = PQ.
  • Draw DE parallel to BC, meeting AC at E.
  • By BPT, AD/AB = AE/AC = DE/BC. Since AD=PQ and AB=k·PQ, we get AE/AC = 1/k and DE = QR.
  • Also AE = AC/k, so AE/AC = 1/k. Thus ΔADE has sides PQ, QR, PR.
  • By SSS congruence, ΔADE ≅ ΔPQR.
  • Since DE || BC, ΔADE ~ ΔABC. Therefore, ΔABC ~ ΔPQR.

Worked example

In ΔABC and ΔDEF, AB = 4 cm, BC = 6 cm, AC = 5 cm, DE = 8 cm, EF = 12 cm, DF = 10 cm. Which of the following is true?

AB/DE = 4/8 = 1/2, BC/EF = 6/12 = 1/2, AC/DF = 5/10 = 1/2. Since all three pairs of corresponding sides are proportional, ΔABC ~ ΔDEF by SSS similarity.

  • ΔABC ~ ΔDEF — correct
  • ΔABC ~ ΔEDF
  • ΔABC ~ ΔDFE
  • The triangles are not similar
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