In triangle ABC, if AL is perpendicular to BC and AM is the bisector of angle A. Show that angle LAM= 1/2 ( angle B - angle C ).
Answers
Given :-
- In triangle ABC, AL is perpendicular to BC.
- AM is the bisector of ∠ A.
To show :-
- ∠LAM = (1/2) [ ∠C - ∠B ] .
Solution :-
from image , we have ,
in Right ∆ALC,
→ ∠ALC + ∠C + ∠CAL = 180° (Angle sum Property.)
→ 90° + ∠C + ∠CAL = 180°
→ ∠C + ∠CAL = 180° - 90°
→ ∠C + ∠CAL = 90°
→ ∠C = (90° - ∠CAL) ------------ Eqn.(1)
Similarly,
in Right ∆ALB,
→ ∠ALB + ∠B + ∠LAB = 180° (Angle sum Property.)
→ 90° + ∠B + ∠LAB = 180°
→ ∠B + ∠LAB = 180° - 90°
→ ∠B + ∠LAB = 90°
→ ∠B = (90° - ∠LAB) ------------ Eqn.(2)
Subtracting Eqn.(2) from Eqn.(1) now,
→ ∠C - ∠B = (90° - ∠CAL) - (90° - ∠LAB)
→ ∠C - ∠B = 90° - 90° + ∠LAB - ∠CAL
→ ∠C - ∠B = ∠LAB - ∠CAL
→ ∠C - ∠B = (∠LAM + ∠MAB) - (∠CAM - ∠LAM)
→ ∠C - ∠B = ∠LAM + ∠LAM + ∠MAB - ∠CAM .
Now, since AM is angle bisector,
=> ∠MAB = ∠CAM
=> ∠MAB - ∠CAM = 0
Therefore,
→ ∠C - ∠B = 2∠LAM
→ ∠LAM = (1/2)[ ∠C - ∠B ] . (Proved).
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