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.

Since LM = AM, point M must be on the perpendicular bisector of AL. Since AM = BM, BL must be perpendicular to AL. This makes ∆ALC a right triangle with hypotenuse AC twice the length of side AL. Hence ∠LAC = ∠LAB = 60°, and AL is angle bisector, median, and altitude.

ΔABC is isosceles with ∠A = 120°, and ∠B = ∠C = 30°.

55? i think maybe

step-by-step explanation:

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3