# How to use congruence and angle bisector theorem for this geometry question involving isosceles triangle?

Here is the diagram for the question: In triangle $\triangle{ADC}$, $|AB|=|AC|=8$ and $|NC|=6$. $AN$ is the bisector of $\measuredangle{BAC}$. $\measuredangle{ADC}=2\cdot\measuredangle{BCD}$. $|DN|=x$. Find $x$.

I determine that $|BE|=|CE|$. I call the angle $\measuredangle{BCD}=\alpha$, $\measuredangle{ADC}=2\alpha$ and $\measuredangle{ABC}=\measuredangle{ACB}=3\alpha$. According to angle bisector theorem i can write:

\begin{align} \frac{8+BD}{x}=\frac{8}{6}=\frac{4}{3} \\ 24+3BD=4x \end{align}

After that i couldn't continue. Solving the question by using congruence, circles, angle bisector theorem, angle transportations, Sinus and Cosinus theorems are more appreciated.

• How did you get that $\angle BCD=\alpha$, and $\angle ADC=2\alpha$? Apr 29, 2017 at 8:08
• On your figure, it seems that $\measuredangle{ADC}\approx\cdot\measuredangle{BCD}$ instead of $\measuredangle{ADC}=2\cdot\measuredangle{BCD}$. I know that we can have a good reasoning on an erroneous figure but I wanted you to say that the elements given in your questions are all correct. Apr 29, 2017 at 8:08
• @TobyMak it has already given, look at the question. I just assigned a varible for it.
– user373239
Apr 29, 2017 at 8:12
• How do you know that $\angle ADC$ is twice as large as $\alpha$, and $\angle ABC$ and $\angle ACB$ are 3 times as large as $\alpha$? Apr 29, 2017 at 8:15
• I transported it. Draw a line parallel to $BC$ and passes through $D$. Observe that $\measuredangle{ABC}$ is the sum of two angles. $\measuredangle{ABC}=\measuredangle{ACB}$ because $ABC$ is a isosceles triangle.
– user373239
Apr 29, 2017 at 8:20

Join $BN$. Note that $\triangle NEC\cong\triangle NEB$ and hence $\angle NBC=\alpha$. So $\angle BND=\angle NBC+\angle NCB=2\alpha=\angle NDB$. So $\triangle BND$ is isosceles with $BD=BN$. Therefore, $BD=BN=CN$.
$$\frac{4x-24}{3}=6$$.
Thus $x=10.5$.
• There's a little typo. It should be $BD=BN$ i guess.