Would my answer be correct?

Question

Triangles ABD and CBD are shown.

If m∠ADB = 110°, what is the relationship between AB and BC?

 AB < BC 

AB > BC

 AB = BC 

AB + BC < AC

I say option 3 but I am not sure please correct me if I am wrong

Answer 1

AB > BC

from the drawing, if to scale, AD =DC, D divides AC in half. AC/2 = AD = DC

ADB = 110 degrees

BDC =180-110 = 70 degrees

in triangles the side opposite the larger angle is > the side opposite a smaller angle

AB is opposite 110

BC is opposite 70

110>70, so

AB >BC

let AD=DC = 10 and let BD = 10

use the law of cosines

AB^2 = 10^2 +10^2 -2(10)(10)Cos110

= 200-200Cos110

= 200+200Cos20=about 388

AB =sqr388 = about 19.7

BC^2 = 10^2 +10^2 -2(10)(10)Cos70

= 200-200Cos70

=about 131.5959713

BC = sqr131.596 = about 11.5

19.7>11.5

AB>BC

2nd option is correct

Answer 2

If two sides of one triangle

are congruent to two sides

of another triangle and

the included angles are not

congruent, then the longer

third side is across from the

larger included angle.

That is called the hinge theorem.

In this problem segment BD is a part of both triangles, so we have two sides of one triangle congruent to two sides of a second triangle. The triangles are ADB and CBD. Since angle ADB measures 110, we have angle CDB = 70 (linear pair supplementary). Since by the Hinge Theorem the longer side is opposite the larger angle, AB is greater than BC.