Given =
AL is an angle bisector (L∈ BC ).
Point M∈ AB
LM=AM=BM
AC=2AL
Find out angles in △ABC.
To proof
As given LM =AM =BM & AC=2AL
AL is an angle bisector i.e AL divide the A in two equal parts.
let ∠ BAL = y ,∠ CAL = y
inΔ MAL
ML = AM (given)
if the two sides of the triangle are equal their opposite angle are also equal.
i.e ∠ MLA =∠ MAL = y
similarly in Δ MBL
MB = ML
if the two sides of the triangle are equal their opposite angle are also equal.
i.e ∠ MBL = ∠MLB = x
thus ∠ ALB = x+y
Now in ΔABL
We have
x + x + y + y = 180
x + y = 90
thus ∠ ALB = 90°
thus also∠ ALC = 90°
As given AC=2AL
Formula
[tex]\sin C = \frac{perpendicular}{hypotenuse}[/tex]
[tex]\sin C = \frac{AL}{AC}[/tex]
[tex]\sin C=\frac{AL}{2AL}[/tex]
[tex]sin C=\frac{1}{2}[/tex]
sin C = 30°
Thus in ΔALC
∠A +∠ L +∠ C = 180 ° ( Proof above ∠ ALC = 90)
∠A = 60°
Thus ∠LAC = ∠LAB = 60°
In ΔALB
∠A +∠ L + ∠B = 180°
∠B = 180 -90 -60
∠B = 30°
Thus
∠ABL = 30°
Hence all the angle in theΔ ABC are
∠ABC = 30°
∠BAC =120°
∠ACB = 30 °
Hence proved
