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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.

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).

## If two triangles have the three sides of the one equal to the three sides of the other, each to each, then they are equal in all respects.

If two triangles have the three sides of the one equal to the three sides ...