In △abc al is an angle bisector (l∈ bc ). point m∈ ab so that lm=am=bm. find the angles in △abc, if ac=2al.

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

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