In the given figure, side AB of △ABC is produced to D such that BD = BC. If ∠A = 60° and ∠B = 50°, prove that :

(i) AD > CD

(ii) AD > AC

We know that,

Sum of angles of triangle = 180°

∴ ∠A + ∠B + ∠C = 180°

⇒ 60° + 50° + ∠C = 180°

⇒ 110° + ∠C = 180°

⇒ ∠C = 180° - 110°

⇒ ∠C = 70°.

From figure,

⇒ ∠CBD + ∠CBA = 180° (linear pair)

⇒ ∠CBD + 50° = 180°

⇒ ∠CBD = 180° - 50°

⇒ ∠CBD = 130°

In △BDC,

BD = BC

∴ ∠BDC = ∠BCD = x (let)

Sum of angles of triangle = 180°

∴ ∠BDC + ∠BCD + ∠CBD = 180°

⇒ x + x + 130° = 180°

⇒ 2x = 180° - 130°

⇒ 2x = 50°

⇒ x = $\dfrac{50°}{2}$

⇒ x = 25°

∴ ∠BDC = ∠BCD = 25°

From figure,

⇒ ∠ACD = ∠C + ∠BCD = 70° + 25° = 95°

(i) In △ADC,

We know that side opposite to the greatest angle is greatest side.

Since, ∠ACD is greatest,

∴ AD is longest side of the triangle.

⇒ AD > CD

Hence, proved that AD > CD.

(ii) Since, AD is greatest side of triangle ADC,

∴ AD > AC

Hence, proved that AD > AC.