Find the measure of a side and an angle.

Question

In the figure, $BG=10$, $AG=13$, $DC=12$, and $m\angle DBC=39^\circ$.

Given that $AB=BC$, find $AD$ and $m\angle ABC$.

Here is the figure:

I am inclined to say that since $\overline{AB}\simeq \overline{BC}$, both triangles share side $\overline{BD}$, and they also have a $90^\circ$ angle in common, then $AD=DC$ and $m\angle ABD=m\angle DBC=39^\circ$. However, I am not making the connection of exactly why my conclusion is true.

How can I show this is true without using trigonometry?

Thank you!

Answer 1

Since triangle $ABC$ is isosceles, with $AB=BC$,

1. To find $\angle ABC$, notice that $\angle BCD=51^0=\angle DAB$, hence $\angle ABD=39^0$, and $\angle ABC=39^o+39^o=78^o $.
2. To find $AD$, since ABC is an isosceles triangle, and since $\angle ABD=\angle DBC$, the line $BD$ divides $AC$ into two equal parts $AD=DC$. Since $DC=12$, hence $AD=12$.

Answer 2

You can use the law of sines with the given angle and length of $DC$ to find the length $BD$, then since you're given $BG$, you can find $GD$, and via pythagorean theorem, find $AD$. (Is it exactly 12 as you predicted?)

Answer 3

You can use the RHS condition for the congruence of two right-angled triangles:

https://en.wikipedia.org/wiki/Congruence_(geometry)#Congruence_of_triangles

It's also called "hypotenuse-leg test (HL)":

http://www.mathopenref.com/congruenthl.html

You then have $BCD\cong BAD$ and that's all.