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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. We have to prove that ∠BDC = 90º+½∠A.
Proof: In the ∆ABC,
∠A+∠B+∠C=180°
⇒½∠A+½∠B+½∠C=½×180° [Multiplying ½ in both side] ⇒½∠B+½∠C=½×180°˗½∠A ——————–(1)
Again, in the ∆BDC,
∠BDC+∠DBC+∠DCD=180°
⇒∠BDC+½∠B+½∠C=180°
⇒ ∠BDC+½×180° ˗ ½∠A=180°
⇒∠BDC=180°-90° + ½∠A
⇒∠BDC = 90° + ½∠A (proved).

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