Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a problem understanding the proof.

Given an acute angle $A$. Choose an arbitrary point $P$ from the bisector of $A$ and another point $B$ from the side of angle $A$. Draw a line $l$ going through points $P$ and $B$. Now $l$ intersects another side of the angle $A$ at point $C$. Prove that no matter how we choose $B$ and keep $P$ fixed, the expression $\frac{1}{\mid AB\mid}+\frac{1}{\mid AC\mid}$ is constant.


Let $\angle A=2\theta$, $\angle ACB = \phi$, and $PD$ is perpendicular to $AC$. Because $AP$ bisects $\angle A$ we have ${AC\over AB}={PC\over PB}$. Thus:

${1\over AB}+{1\over AC}={1\over AB}+{PB\over AB\cdot PC}={BC\over AB}\cdot{1\over PC}={\sin 2\theta\over {PC\cdot\sin\phi}}={\sin 2\theta\over PD}={\sin 2\theta\over {PA\cdot\sin \theta}}={2\cos\theta\over PA}={\rm constant,}$

because both $\theta$ and $PA$ are given.

Why do we have that ${1\over AB}+{PB\over AB\cdot PC}={BC\over AB}\cdot{1\over PC}$ and is that proof correct?

share|cite|improve this question
up vote 1 down vote accepted

In virtue of the angle bisector theorem, $$\frac{PB}{PC}=\frac{AB}{AC},$$ then: $$\frac{1}{AB}+\frac{1}{AC}=\frac{1}{AB}+\frac{PB}{AB\cdot PC}=\frac{1}{AB}\left(1+\frac{PB}{PC}\right)=\frac{1}{AB}\cdot\frac{BC}{PC}.$$

It is interesting to note that, if we take $B'$ as the symmetric of $B$ wrt to $AP$, $C'$ as the symmetric of $C$ wrt to $AP$, the diagonals of the isosceles trapezoid $BB'CC'$ meet in $P$. If we take the perpendicular to $AP$ through $P$, the length of the segment cut by the angle is equal to the harmonic mean of $BB'$ and $CC'$, so the result follows.

Another way is to consider the angle as a skew reference system and the point $P$ as $(1,1)$. In this setting, all the lines throug $P$ have equation $ax+by=a+b$, so they intersect the coordinate axis in $y_0=\frac{a+b}{b}$ and $x_0=\frac{a+b}{a}$: this gives $\frac{1}{x_0}+\frac{1}{y_0}=1$, QED.

share|cite|improve this answer
@student: if the answer is useful, you should consider upvoting. – robjohn Nov 16 '12 at 15:28
I haven't registered to the forum. – student Nov 16 '12 at 17:59

Area APB + Area APC = Area ABC $$\frac{1}{2}PB\sin(θ)+\frac{1}{2}PC\sin(θ) = \frac{1}{2}BC\sin(2θ)$$ $$\sin(2θ)= 2\sin(θ)cos(θ) \\ \Rightarrow PB+PC= 2BC\cos(θ)$$ (divide by $BCP$) $$\Rightarrow \frac{1}{C}+\frac{1}{B} =2\frac{\cos(θ)}{P}$$

share|cite|improve this answer
Please try to learn some Latex, it makes proofs much easier to follow. – Simon Hayward Nov 17 '12 at 13:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.