Question

The given figure shows a circle with centre O and radius 4 cm circumscribed by \(\triangle ABC\). BC touches the circle at D such that BD = 6 cm, DC = 10 cm. Find the length of AE.

Answer 1

Find AE in Triangle with Incircle

Given:

• Radius \( r = OE = OF = OD = 4 \, \text{cm} \)
• BD = 6 cm, DC = 10 cm

Step 1: Use Tangents from External Points

Tangents from the same external point to a circle are equal in length.

• From point B: \( BF = BD = 6 \, \text{cm} \)
• From point C: \( CE = DC = 10 \, \text{cm} \)
• From point A: Let \( AE = AF = x \, \text{cm} \)

Step 2: Express Side Lengths

\[ AB = AF + FB = x + 6 \quad AC = AE + EC = x + 10 \quad BC = BD + DC = 6 + 10 = 16 \]

Step 3: Area Method 1 (Using Triangles)

• Area of \( \triangle OAB = \frac{1}{2} \cdot AB \cdot OF = \frac{1}{2}(x + 6)(4) = 2(x + 6) \)
• Area of \( \triangle OBC = \frac{1}{2} \cdot BC \cdot OD = \frac{1}{2}(16)(4) = 32 \)
• Area of \( \triangle OCA = \frac{1}{2} \cdot AC \cdot OE = \frac{1}{2}(x + 10)(4) = 2(x + 10) \)

Total Area: \[ A = 2(x + 6) + 32 + 2(x + 10) = 4x + 64 \]

Step 4: Area Method 2 (Heron’s Formula)

Semi-perimeter: \[ s = \frac{AB + BC + AC}{2} = \frac{(x + 6) + 16 + (x + 10)}{2} = \frac{2x + 32}{2} = x + 16 \] Now: \[ s - AB = x + 16 - (x + 6) = 10 \\ s - BC = x + 16 - 16 = x \\ s - AC = x + 16 - (x + 10) = 6 \] So, \[ A = \sqrt{(x + 16)(10)(x)(6)} = \sqrt{60x(x + 16)} \]

Step 5: Equating the Areas

\[ 4x + 64 = \sqrt{60x(x + 16)} \] Square both sides: \[ (4x + 64)^2 = 60x(x + 16) \] Expand both sides: \[ 16x^2 + 512x + 4096 = 60x^2 + 960x \] Simplifying: \[ 0 = 44x^2 + 448x - 4096 \Rightarrow x = \frac{256}{44} = \frac{64}{11} \] Therefore, \[ \boxed{AE = \frac{64}{11} \, \text{cm}} \]