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Home / Geometry / In the ∆ABC, the internal bisector of the ∠B and ∠C intersects at the point D. Prove that ∠BDC = 90°-½∠A.

In the ∆ABC, the internal bisector of the ∠B and ∠C intersects at the point D. Prove that ∠BDC = 90°-½∠A.

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.

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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=1800

In the triangle ∆BDC,

∠BDC+∠BCD+∠DBC=1800

But ∠BCD = ½∠BCF = ½(∠A+∠B)

and ∠DBC = ½∠CBF = ½(∠A+∠C)

∴∠BDC =1800– ½(∠A+∠A+∠B +∠C)

⇒∠BDC =1800– ½(∠A+1800) [∠A+∠B+∠C=1800]

⇒∠BDC=1800-900 – ½∠A

⇒∠BDC = 900 – ½∠A   (proved).

 

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