16. Length of the (interior) angle bisector (10 proofs).

by Virgil Nicula, Apr 20, 2010, 2:55 AM

Sa dau cateva detalii relativ la aceasta problema. Cu multi ani in urma la clasa a IX - a dupa prima saptamana de scoala am dat un test de verificare a cunostintelor

de la gimnaziu prin care "masuram" (din punctul meu de vedere) nivelul de pregatire al elevilor pe care ii "adoptasem" pentru a ramane impreuna pana la terminarea

liceului. Rezultatele de la "treapta I - a" erau semnificative, insa pentru mine nu erau suficient de concludente. Mare mi-a fost surpriza ca la aceasta problema sa gasesc

printre lucrari cateva solutii (unele folosind chiar simple elemente de trigonometrie), care mai de care mai frumoase. Le-am pastrat si recent am dat intamplator peste ele.



PP1. Prove that the length $l_a$ of the interior $A$-angled bisector in $\triangle ABC$ is given by the relation $\boxed{\ l_a=\frac {2bc}{b+c}\cdot\cos\frac A2\ }$ .

Method 1 (Mihai). Let $\left\{\begin{array}{ccc}
D\in (BC) & ; & \angle  DAB\equiv\angle DAC\\\\
E\in (AB) & ; & DE\ \parallel\ AC\end{array}\right\|$ $\implies$ $EA=ED$ and $\frac {AB}{AE}=\frac {CB}{CD}$ $\implies$ $\frac {c}{EA}=\frac a{\frac {ab}{b+c}}\implies$ $\left\{\begin{array}{ccc}
EA & = & \frac {bc}{b+c}\\\\
AD & = & 2\cdot ED\cdot\cos\frac A2\end{array}\right\|$ $\Longrightarrow\ AD=\frac {2bc\cdot\cos\frac A2}{b+c}$ .

Method 2 (Emil). Let $\left\{\begin{array}{ccc}
D\in (BC) & ; & \widehat{DAB}\equiv\widehat{DAC}\\\\
E\in AB & ; & CE\ \parallel\ AD\end{array}\right\|$ . Prove easily that $AE=b$ and $\left\{\begin{array}{ccc}
CE & = & 2b\cdot\cos\frac A2\\\\
\frac {AD}{CE} & = & \frac {c}{b+c}\end{array}\right\|$ $\Longrightarrow\ AD=\frac {2bc\cdot\cos\frac A2}{b+c}$ .

Method 3 (Ioan). Let $\left\{\begin{array}{ccc}
D\in (BC)  & ; & \widehat{DAB}\equiv\widehat{DAC}\\\\
E\in AD & ; & BE\ \parallel\ AC\end{array}\right\|$ . Prove easily that $BE=BA=c$ and $\left\{\begin{array}{ccc}
AE & = & 2c\cdot\cos\frac A2\\\\
\frac {AD}{b}=\frac {DE}{c} & = & \frac {AE}{b+c}\end{array}\right\|$ $\Longrightarrow\ AD=\frac {2bc\cdot\cos\frac A2}{b+c}$ .

Method 4 (Dan). Let $D\in (BC)$ for which $m\left(\widehat{DAB}\right)=m\left(\widehat{DAC}\right)=x$ . I"ll use the relations $2S=bc\cdot\sin A$ and

$\sin 2x=2\sin x\cos x$ . Thus $[ABC]=[DAB]+[DAC]\ \Longrightarrow\ bc\cdot\sin 2x=$ $AD\cdot (b+c)\cdot\sin x\ \Longrightarrow\ AD=$ $\frac {2bc\cos\frac A2}{b+c}$ .



$(\ \Downarrow\ )$ Another proofs for the $\mathrm{IX}^{\mathrm{th}}$ class.

Method 5 (prof. Constantin Mateescu). Suppose w.l.o.g. $b\ge c$ . Denote $D\in (BC)$ , $\widehat{DAB}\equiv\widehat{DAC}$ and $E\in AB$ , $F\in AC$ such that $D\in EF$

and $AD\perp\overline {EDF}$ . Observe that $L\equiv AE=AF=\frac {AD}{\cos\frac A2}$ and $\frac {DB}{c}=\frac {DC}{b}=\frac {a}{b+c}$ . Apply Menelaus' theorem to the transversal $\overline {EDF}/\triangle ABC\ :$

$\frac {EB}{EA}\cdot\frac {FA}{FC}\cdot\frac {DC}{DB}=1$ $\Longrightarrow\ EB\cdot DC=FC\cdot DB\ \Longrightarrow\ b(L-c)=c(b-L)\ \Longrightarrow$ $\ L=\frac {2bc}{b+c}\ \Longrightarrow\ AD=\frac {2bc\cdot\cos\frac A2}{b+c}$ .


Method 6 (prof. Virgil Nicula). Apply an well-known property $:\ \frac {BD}{BC}=\frac {AD}{AC}\cdot\frac {\sin\widehat{BAD}}{\sin\widehat{BAC}}\iff$ $\frac {c}{b+c}=\frac {l_a}{b}\cdot\frac {\sin\frac A2}{\sin A}\iff$ $\frac {c}{b+c}=\frac {l_a}{b}\cdot\frac {1}{2\cos\frac A2}\iff$ $\boxed{l_a=\frac {2bc\cos\frac A2}{b+c}}$ .

Method 7 (prof. Virgil Nicula). Apply the theorem of Sinus in $\triangle ABD\ :\ \frac {BD}{AD}=\frac {\sin \frac A2}{\sin B}\iff$ $AD=\frac {ac\sin B}{(b+c)\sin\frac A2}=$ $\frac {bc\sin A}{(b+c)\sin\frac A2}\implies$ $\boxed{l_a=\frac {2bc\cos\frac A2}{b+c}}$ .

Method 8 (prof. Virgil Nicula). Let circumcircle $w$ of $\triangle ABC$ and $\{A,S\}=AD\cap w$ . Thus, $\triangle ABS\sim\triangle ADC\ \Longrightarrow$ $AD\cdot AS=bc$ . The power of $D$ w.r.t. $w\ \Longrightarrow\ DA\cdot DS=$ $DB\cdot DC$

$\Longrightarrow$ $AD\cdot (AS-AD)=$ $DB\cdot DC\ \Longrightarrow$ $AD^2=$ $bc-DB\cdot DC$ , where $\frac {DB}{c}=\frac {DC}{b}=$ $\frac {a}{b+c}$ $\Longrightarrow$ $AD^2=bc-\frac {a^2bc}{(b+c)^2}$ $\Longrightarrow$ $\boxed {\ AD=\frac {2\sqrt {bcp(p-a)}}{b+c}=\frac {2bc\cdot\cos\frac A2}{b+c}\ }$ .

Method 9 (prof. Virgil Nicula). Let $D\in (BC)$ so that $m(\angle DAB)=m(\angle DAC)=x$ and $\left\|\begin{array}{c}
 X\in AC\ ,\ BX\ \parallel\ AD\\\\
 Y\in AB\ ,\ CY\ \parallel\ AD\end{array}\right\|$ . Thus the trapezoid $XBCY$ is isosceles and

$\frac {BX}{c}=\frac {CY}{b}=2\cdot\cos\frac A2$ . But $\left\|\begin{array}{c}
 \frac {CD}{CB}=\frac {AD}{BX}\\\\
 \frac {BD}{BC}=\frac {AD}{CY}\end{array}\right\|\ \bigoplus\ \Longrightarrow$ $\frac {1}{AD}=\frac {1}{BX}+\frac {1}{CY}\ \Longrightarrow\ \frac 1b+\frac 1c=$ $\frac {2\cdot\cos\frac A2}{AD}\ \Longrightarrow\ AD=$ $\frac {2bc\cdot\cos\frac A2}{b+c}$ . This proof is symetrically in $b$ , $c$ .


Method 10. Apply Stewart's relation $:\ AB^2\cdot DC+AC^2\cdot DB=AD^2\cdot BC+BC\cdot DB\cdot DC\iff$ $c^2\cdot\frac {ab}{b+c}+b^2\cdot\frac {ac}{b+c}=AD^2\cdot a+a\cdot DB\cdot DC\iff$ $\boxed{bc=AD^2+DB\cdot DC}\ .$

Extension. $\triangle ABC\ \ \wedge\ \ D\in (BC)\ \ \wedge\ \ \left\|\ \begin{array}{c}
 m(\angle DAB)=x\\\\
 m(\angle DAC)=y\end{array}\ \right\|$ $\Longrightarrow$ $\boxed {\ \frac {\sin x}{b}\ +\ \frac {\sin y}{c}\ =\ \frac {\sin A}{AD}\ }\ (*)$ .


Proof 1. $\left\|\ \begin{array}{ccc}
 \frac {BD}{BC}=\frac {AD}{AC}\cdot\frac {\sin (\angle BAD)}{\sin (\angle BAC)} & \iff & \frac {c}{b+c}=\frac {AD}{b}\cdot\frac {\sin x}{\sin A}\\\\
 \frac {CD}{CB}=\frac {AD}{AB}\cdot\frac {\sin (\angle CAD)}{\sin (\angle CAB)} & \iff & \frac {b}{b+c}=\frac {AD}{c}\cdot\frac {\sin y}{\sin A}\end{array}\ \right\|\ \bigoplus\ \Longrightarrow\ 1=$ $\frac {AD}{\sin A}\cdot \left(\frac {\sin x}{b}+\frac {\sin y}{c}\right)\ \Longrightarrow\ \frac {\sin x}{b}\ +$ $\ \frac {\sin y}{c}\ =\ \frac {\sin A}{AD}$ .

Proof 2. $[ABC]=[ABD]+[ADC]\Longleftrightarrow bc\cdot \sin A=$ $c\cdot AD\cdot\sin x+b\cdot AD\cdot\sin y\Longleftrightarrow$ $ \frac {\sin x}{b}+\frac {\sin y}{c}=\frac {\sin A}{AD}$ .

Remark. Let $\{\ A\ ,\ E\ \}=AD\ \cap\ w$ , where $w$ is the circumcircle of $\triangle ABC$ . Then $(*)$ is equivalently with $\boxed {\ \frac {BE}{b}\ +\ \frac {CE}{c}\ =\ \frac {a}{AD}\ }$ . If $\triangle ABC$ is acute and $D\in (BC)$ , then:

$1.\ \odot\ AD\perp BC \Longrightarrow\ \frac {\cos B}{b}+\frac {\cos C}{c}=\frac {\sin A}{h_a}$ , where $h_a$ is length of the $A$-altitude $\Longleftrightarrow\ b\cdot\cos C+c\cdot\cos B=a$ .

$2.\ \odot\ O\in AD\Longrightarrow\ \frac {\cos C}{b}+$ $\frac {\cos B}{c}=\frac {\sin A}{AD}$ , where $O$ is the circumcenter of $\triangle ABC\ \Longleftrightarrow\ b\cdot \cos B+c\cdot\cos C=a\cdot\cos (B-C)$ .

Concluzie. A fost odata ca niciodata. Programa actuala este departe de a pregati asemenea elevi decat printr-o pregatire suplimentara prin meditatii, consultatii

si in cadrul cercurilor de matematica care nu sunt realizabile in cadrul scolii decat ori obligati la o "munca patriotica" asidua ori gratuit prin dragostea unor profesori

pentru elevii lor, de a-i pregati pentru a face performanta nu numai in matematica, dar si in viata. In unele tari civilizate lucrurile stau altfel. Sunt profesori care in

exclusivitate si in afara orelor din program ori pregatesc elevii care au un talent deosebit intr-un anumit domeniu de studiu ori se ocupa de elevii care au dificultati

in invatare si rezolvarea temelor pentru a doua zi. Toate acestea se realizeaza pe nivele si in clase special amenajate in cadrul scolii. Si aici ma refer nu numai la

matematica, fizica, chimie, literatura etc, dar si la muzica, pictura, educatie fizica (anumite sporturi individuale sau colective) etc. Va recomand cartea profesorului

Bogdan Enescu despre "Arii". Pentru cei de la gimnaziu sau liceu este o adevarata "bijuterie", foarte utila chiar si pentru probleme ceva mai "delicate".
This post has been edited 96 times. Last edited by Virgil Nicula, Jun 23, 2017, 6:10 AM

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