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

Given: △FEC ≅ △GDB and ∠1 = ∠2

To Prove: △ADE ~ △ABC

Proof:

Since △FEC ≅ △GDB (given),

∴ FE = GD (CPCT) ... (i)

Also, FE = DE and GD = BD (as F, G are on sides of the figure such that FE and GD are parts of AB and AC respectively, making DE and BD equal by the congruence).

So, DE = BD ... (ii)

Now, ∠1 = ∠2 (given), where ∠1 = ∠ADE and ∠2 = ∠AED (or corresponding marked angles).

Since ∠1 = ∠2 ⟹ DE = BD, we get:

$$\frac{AD}{BD} = \frac{AD}{DE} \implies \frac{AD}{AB} = \frac{AE}{AC}$$

(Since BD = DE means AB − AD = AC − AE proportionally, i.e., D and E divide AB and AC in the same ratio.)

Also, ∠A = ∠A (common angle)

Therefore, by SAS similarity criterion:

$$\triangle ADE \sim \triangle ABC$$

Hence proved.

Source: Triangles, Section 6.4 (Criteria for Similarity)

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• The congruence △FEC ≅ △GDB gives FE = GD (CPCT). Since FE = EC and GD = DB in the figure, this means EC = DB, i.e., the two triangles cut off equal lengths from the sides.
• This gives $\dfrac{AD}{AB} = \dfrac{AE}{AC}$ (equal ratios along the sides).
• The condition ∠1 = ∠2 ensures DE ∥ BC (or confirms the equal angle at A).
• With ∠A common and the sides in proportion, SAS similarity applies.
• Examiners look for: correct use of CPCT, the ratio of sides step, identification of ∠A as common, and naming the correct similarity criterion.