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Use the two established similarities to prove the product identity.

Read the idea, work independently, then explain what changed.

TOPIC 01

2025 JM01

Ratios must follow the vertex correspondence, not visual side lengths.

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Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Standard#Your turn

Use the two established similarities to prove the product identity.

DG⋅DF=DB⋅EFDG\cdot DF=DB\cdot EF
Original schematic for the triangle-similarity proofABCDEFG

Official paper · jm01-2025 · II.4(c) · PDF 4

Official original and suggested answers ↗ · Suggested answer PDF page 7

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express DE² using each similarity.
Hint 2
Equate the two products.
Worked solution
  1. The first similarity gives a mean-proportional relation.

    DEDF=DGDE  ⟹  DE2=DG⋅DF\frac{DE}{DF}=\frac{DG}{DE}\implies DE^2=DG\cdot DF
  2. The second similarity expresses the same square.

    DEDB=EFDE  ⟹  DE2=DB⋅EF\frac{DE}{DB}=\frac{EF}{DE}\implies DE^2=DB\cdot EF
  3. Both expressions equal DE².

    DG⋅DF=DB⋅EFDG\cdot DF=DB\cdot EF

DG·DF=DB·EF.

Checks and common pitfalls: Ratios must follow the vertex correspondence, not visual side lengths.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • The first similarity gives a mean-proportional relation.
    DEDF=DGDE  ⟹  DE2=DG⋅DF\frac{DE}{DF}=\frac{DG}{DE}\implies DE^2=DG\cdot DF
  • The second similarity expresses the same square.
    DEDB=EFDE  ⟹  DE2=DB⋅EF\frac{DE}{DB}=\frac{EF}{DE}\implies DE^2=DB\cdot EF
  • Both expressions equal DE².
    DG⋅DF=DB⋅EFDG\cdot DF=DB\cdot EF

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Curriculum and source notes ↗