Corresponding sides connect corresponding vertices. If A matches F and B matches D, side AB matches side FD. This remains true if one triangle is upside down or reflected. The endpoints define the pairing more reliably than descriptions such as top, left or bottom.

An equivalent check uses opposite angles. Side BC is opposite A, and DE is opposite F, so those sides correspond when A matches F. Use the actual marked equalities or a supplied similarity statement. A sketch’s resemblance alone is not a proof of the required relationship.

With A ↔ F, B ↔ D and C ↔ E, write triangle ABC similar to triangle FDE. The order records the correspondence. It does not say that A automatically matches D merely because D is the first alphabetic letter in the other triangle’s name.

If the problem instead states triangle ABC similar to triangle PQR, the matching is A ↔ P, B ↔ Q and C ↔ R. Read the supplied order carefully. Rewriting the same triangle’s letters in a different order can change which correspondence a similarity statement asserts.

State the geometry completely: angle A and angle F are right angles, angle B equals angle D, and hence angle C equals angle E. In triangle ABC, AB = 3, AC = 4 and BC = 5 cm. The hypotenuse BC matches DE, while the two legs match FD and FE respectively.

If FD = 6 cm, the length scale factor from ABC to FDE is 6/3 = 2. Therefore FE = 2 × 4 = 8 cm and DE = 2 × 5 = 10 cm. The ratios 6/3, 8/4 and 10/5 agree. Each uses image length divided by its matched original length.

A rotation can move the original bottom side to a vertical position in the image. A reflection can reverse the left-right appearance while keeping equal corresponding angles and proportional lengths. Choosing sides by where they sit can therefore create a false scale factor before any algebra begins.

One useful exercise rotates a labelled sketch while leaving its vertex letters attached. Ask your child whether the correspondence changed. It did not. The aim is not to guess how the diagram was moved, but to use the relationships that remain established despite the movement.

Say the direction aloud. From a triangle with corresponding length three to one with length six, the factor is two. From six back to three, it is one half. Both are correct in their stated directions. Mixing one forward ratio with one reverse ratio produces an inconsistent proportion.

Write a short heading such as ‘large/small’ or ‘image/original’ above every ratio. For the main example, 6/3 = 8/4 is consistent; 6/3 = 4/8 is not. If an answer makes a supposedly larger corresponding side smaller, recheck both the matching and the ratio direction.

Two equal corresponding angle pairs establish triangle similarity because the third pair then agrees. Other valid triangle criteria use proportional corresponding sides, or two proportional side pairs with the included angles equal. Read which evidence the question actually supplies.

Two arbitrary side ratios alone are not a general similarity proof without the required further condition. Nor does one right angle in each triangle prove that all right triangles are similar. In our main example the additional marked equality B = D is essential to the angle-based argument.

Question 1: triangle ABC is similar to triangle PQR; which side matches AC? Question 2: AB = 4 cm, BC = 6 cm, AC = 8 cm and PQ = 6 cm; find QR and PR. Question 3: what is the reverse factor from PQR to ABC? Question 4: does rotating one of these triangles change the correspondence?

Answers: 1) PR. 2) The forward factor is 6/4 = 1.5, giving QR = 9 cm and PR = 12 cm. 3) Two thirds. 4) No, provided the labels remain attached to their vertices. Ask for the vertex matching before the side calculation.

Bring the complete diagram with its angle markings and the first ratio your child wrote. A useful note is ‘pairs sides by page position rather than corresponding vertices’. In a small-group 3-pax lesson, students can annotate rotated copies and explain why the pairing remains unchanged.

For parents considering Secondary 3 Mathematics tuition in Punggol, the next-question routine is to establish similarity, record vertex correspondence, match endpoints and state the factor’s direction. A clear ratio then follows from the geometry. This supports later area and volume work without losing the original length relationships.