Here is the diagram for the question:

enter image description here

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.

  • $\begingroup$ How did you get that $\angle BCD=\alpha$, and $\angle ADC=2\alpha$? $\endgroup$
    – Toby Mak
    Apr 29, 2017 at 8:08
  • $\begingroup$ 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. $\endgroup$
    – Jean Marie
    Apr 29, 2017 at 8:08
  • $\begingroup$ @TobyMak it has already given, look at the question. I just assigned a varible for it. $\endgroup$
    – user373239
    Apr 29, 2017 at 8:12
  • $\begingroup$ 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 $\alpha$? $\endgroup$
    – Toby Mak
    Apr 29, 2017 at 8:15
  • $\begingroup$ 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. $\endgroup$
    – user373239
    Apr 29, 2017 at 8:20

1 Answer 1


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, $BD=BN=CN$.


Thus $x=10.5$.

  • $\begingroup$ There's a little typo. It should be $BD=BN$ i guess. $\endgroup$
    – user373239
    Apr 29, 2017 at 9:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy