In ΔABC,  AB = AC = 12 cm, BC = 5 cm and D is a point on AC such that DB = BC. What is the measure of CD?

Given:

In triangle ABC, AB = AC = 12 cm.

BC = 5 cm.

D is a point on AC such that DB = BC.

Formula used:

Cosine Rule in a triangle: In any triangle with sides a, b, c and angle C opposite to side c,

c² = a² + b² - 2ab cos(C).

Calculation:

In △ABC, AB = 12 cm, AC = 12 cm, BC = 5 cm.

Since AB = AC, △ABC is an isosceles triangle.

Let's find cos(∠C) in △ABC using the Cosine Rule:

AB² = BC² + AC² - 2 × BC × AC × cos(∠C)

12² = 5² + 12² - 2 × 5 × 12 × cos(∠C)

144 = 25 + 144 - 120 × cos(∠C)

0 = 25 - 120 × cos(∠C)

120 × cos(∠C) = 25

cos(∠C) = 25 / 120

cos(∠C) = 5 / 24

Now consider △DBC.

We are given DB = BC. Since BC = 5 cm, then DB = 5 cm.

We know ∠C is common to both triangles. Let CD = x cm.

Apply the Cosine Rule in △DBC to find CD:

DB² = BC² + CD² - 2 × BC × CD × cos(∠C)

5² = 5² + x² - 2 × 5 × x × (5 / 24)

25 = 25 + x² - (50x / 24)

0 = x² - (25x / 12)

Since x = CD cannot be 0 (as D is a point on AC), we can divide by x:

0 = x - (25 / 12)

x = 25 / 12 cm

∴ The correct answer is option 2.