Say triangle $ABC$ has a right angle at $B$, where $AB=1$ and $BC=2$. The bisector of angle $BAC$ meets $BC$ at $D$. Find a length of $BD$.

I seem to have forgot the property that would help me solve this problem. Can someone refresh my memory?

  • 1
    $\begingroup$ $BD = AB*tan(BAC/2)$ $\endgroup$
    – Coolwater
    Sep 21, 2017 at 17:28

2 Answers 2


let $$BD=x,DC=y$$ then we get $$\frac{x}{y}=\frac{1}{\sqrt{5}}$$ with $$DC=2-x$$ we obtain $$\frac{x}{2-x}=\frac{1}{\sqrt{5}}$$ $$\sqrt{5}x=2-x$$ therefore $$x=\frac{2}{1+\sqrt{5}}$$ the Theorem we used: every bisectrica divides the opposite side in the Proportion of adjacent sides

  • $\begingroup$ Where did you get $x/y$=1/sqrt(5)? $\endgroup$
    – Gerard L.
    Sep 21, 2017 at 17:59
  • $\begingroup$ $$\sqrt{5}$$ is the length of the Hypotenuse and the Quotient Comes from the Theorem above $\endgroup$ Sep 21, 2017 at 18:02

Let $M\in BC$, $K\in AB$ such that $BMDK$ is a square.

Also let $BD=x$.


$$\frac{DK}{BC}=\frac{AK}{AB}$$ or $$\frac{\frac{x}{\sqrt2}}{2}=\frac{1-\frac{x}{\sqrt2}}{1},$$

which gives $$BD=\frac{2\sqrt2}{3}.$$

Another way.

Since $$\frac{CD}{DA}=\frac{BC}{AB},$$ $$AC=\sqrt5$$ and $$BD^2=BC\cdot AB-CD\cdot AD,$$ we obtain: $$BD=\sqrt{2\cdot1-\frac{2\sqrt5}{3}\cdot\frac{\sqrt5}{3}}=\frac{2\sqrt2}{3}.$$


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