Mathematics — Standard (041) — CBSE Class 10 board question

Similarity Proof:

Since AB ∥ DE and BC ∥ EF (as EB = CF = 5 cm, so EF = EC + CF = EC + 5, and BC = 10 cm, so EC = BC − CF... )

Actually, since EB = CF = 5 cm:

• EC = BC − (something)... Let us use the parallel lines directly.

Since AB ∥ DE and AC ∥ DF, quadrilateral ABFD (or the figure formed) has both pairs of opposite sides parallel, so ∠ABC = ∠DEF and ∠BAC = ∠EDF.
By AA similarity criterion, △ABC ∼ △DEF.

Finding DE:

EF = EB + BC + CF... No — from the figure, B and E are on the same line with EB = 5 cm, so:

$$EF = EB + BC - \text{...}$$

Since CF = 5 cm and EB = 5 cm, and BC = 10 cm:
$$EF = BC + EB + CF = 10 + 5 + 5 = 20 \text{ cm}$$

Since △ABC ∼ △DEF:
$$\frac{AB}{DE} = \frac{BC}{EF}$$
$$\frac{7}{DE} = \frac{10}{20} = \frac{1}{2}$$
$$DE = 14 \text{ cm}$$

Source: Chapter 6, Section 6.4 (AA Similarity Criterion)

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• Similarity: AB ∥ DE gives a pair of equal corresponding angles, and AC ∥ DF gives another — AA criterion is sufficient.
• EF: Points are collinear with B between E and C, and C between B and F, so EF = EB + BC + CF = 5 + 10 + 5 = 20 cm.
• Use the ratio of corresponding sides from the similarity to find DE.
• Examiner expects the similarity statement written symbolically, the criterion named, and the ratio set up clearly.