Solution:

General enunciation: In the ∆ABC, the internal bisector of the ∠B and ∠C intersects at the point D. We have to prove that ∠BDC = 90°-½∠A.

Particular enunciation: In the ∆ABC, the internal bisector of the ∠B and ∠C is BD and CD intersects at the point D. °

Proof: In the ∆ABC,

∠A+∠B+∠C=180^{0}

In the triangle ∆BDC,

∠BDC+∠BCD+∠DBC=180^{0}

But ∠BCD = ½∠BCF = ½(∠A+∠B)

and ∠DBC = ½∠CBF = ½(∠A+∠C)

∴∠BDC =180^{0}– ½(∠A+∠A+∠B +∠C)

⇒∠BDC =180^{0}– ½(∠A+180^{0}) [∠A+∠B+∠C=180^{0}]

⇒∠BDC=180^{0}-90^{0 }– ½∠A

⇒∠BDC = 90^{0 }– ½∠A __(proved).__