iff condition on AI+CD=AC where AD is the angle bisector

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iff condition on AI+CD=AC where AD is the angle bisector

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#1 h Mar 6, 2012, 7:14 PM • 2 Y

Y by Adventure10, Mango247

In a triangle $ABC$ , $I$ is the incentre and $D$ the intersection point of $AI$ and $BC$ . Show that $AI+CD=AC$ if and only if $\angle B=60^{\circ}+\frac{_1}{^3}\angle C$ .

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#2 h Mar 6, 2012, 9:00 PM • 2 Y

Y by Adventure10 and 1 other user

Note that $\angle B=60^{\circ}+\frac{_{1}}{^{3}}\angle C\iff 180-\angle A-\angle C=60^{\circ}+\frac{_{1}}{^{3}}\angle C\iff 90^{\circ}=\frac{_3}{^4}\angle A+\angle C$ . But $90^{\circ}=\frac{_1}{^2}\angle A+\frac{_1}{^2}\angle B+\frac{_1}{^2}\angle C$ so we get $\angle B=60^{\circ}+\frac{_{1}}{^{3}}\angle C\iff\frac{_1}{^2}\angle B=\frac{_1}{^4}\angle A+\frac{_1}{^2}\angle C$ .

First suppose $AI+CD=AC$ and let $E$ be the point on $AC$ such that $CE=CD$ . Clearly we also have $AI=AE$ . Now $\triangle AIE$ is $A$ -isosceles so $\angle AEI=\angle AIE=90^{\circ}-\frac{1}{4}A$ . Note also $\angle CED=\angle CDE=90^{\circ}-\frac{1}{2}C$ . Therefore $\angle IED=180^{\circ}-\angle AEI-\angle CED=\frac{C}{2}+\frac{A}{4}$ . Note that $\angle IED+\angle IDE=\angle AIE$ which simplifies to $\angle EDI=\frac{B}{2}$ .

Thus $\angle B=60^{\circ}+\frac{_{1}}{^{3}}\angle C\iff\frac{_1}{^2}\angle B=\frac{_1}{^4}\angle A+\frac{_1}{^2}\angle C\iff \angle IED=\angle IDE$ i.e. $\triangle IDE$ is isosceles.

Let $BI$ meet $DE$ at $F$ . Now we know that $\angle ADF=\frac{B}{2}=\angle ABF$ so $ABDF$ is cyclic. Hence $\angle EFI=\angle DFB=\angle DAB=\angle EAI$ hence $AIEF$ is also cyclic. Then $\angle AEI=\angle AFI=\angle AFB=\angle ADB$ . It follows that $\angle CEI=\angle CDI$ which implies $\angle CEI=\angle EDI$ hence $\triangle IDE$ is isosceles as required.

Next assume we know that $\angle B=60^{\circ}+\frac{_{1}}{^{3}}\angle C$ , we need to prove that $AI=AE$ . Replacing $C$ with $180^{\circ}-A-B$ in the angle condition and simplifying gives $\frac{_1}{^2}A+2B=180^{\circ}$ . Hence, by considering $\triangle ADB$ , it follows that $\angle ADB=B$ . Note that $\triangle CDE$ is isosceles so $\angle EDC=90^{\circ}-\frac{C}{2}$ . This means $\angle EDI=90^{\circ}+\frac{C}{2}-B$ . However this expression is equal to $\frac{B}{2}$ by rearranging the condition $\angle B=60^{\circ}+\frac{_{1}}{^{3}}\angle C$ . Therefore $\angle FDA=\frac{B}{2}=\angle ABF$ which implies $ABDF$ is cyclic. Then $\angle EFI=\angle DFB=\angle DAB=\frac{A}{2}=\angle EAI$ so $AIEF$ is cyclic. Note that $\angle FAD=\angle FBD=\angle ABF=\angle ADF$ so $\triangle FDA$ is $F$ -isosceles. We've mentioned before $\angle EDI=\frac{B}{2}$ so $\angle DFA=180^{\circ}-B$ . SInce $AIEF$ is cyclic, this means $\angle AIE=180^{\circ}-\angle DFA=B=90^{\circ}-\frac{A}{4}$ (as we know $\frac{_1}{^2}A+2B=180^{\circ}$ , so halving and subtracting the $\frac{_1}{^4}A$ gets you the result). But then $\angle AEI=180^{\circ}-\angle AIE-\angle EIA=180^{\circ}-\left(90^{\circ}-\frac{A}{4}\right)-\frac{A}{2}=90^{\circ}-\frac{A}{4}$ . Hence $\angle AEI=\angle AIE$ and so $AE=AI$ which implies $AC=AI+CD$ .

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#3 h Mar 31, 2013, 1:28 PM • 2 Y

Y by Adventure10, Mango247

$\angle B=60^{\circ}+\frac{_1}{^3}\angle C$ implies that $AI+CD=AC$ is equivalent with $a^{2}b=(b-c)(b^{2}+c^{2})$
Can someone helps me how to finish this?

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#4 h Apr 1, 2013, 7:06 PM • 2 Y

Y by Adventure10 and 1 other user

if $\angle B=60 + \angle C/3$ we have $AB=AD$ now extend $AD$ to meet the circumcircle at $S$ . now since $ASC\sim ABD$ we have $AC=AS=AI+IS=AI+CS=AI+CD$ since $\angle CDS=\angle ADB=\angle ABD=\angle CSD$ .
if $AI+CD=AC$ find $L$ on $AC$ such that $AL=AI$ now we have $CL=CD$ but since $\angle ALI=90+\angle A/4$ and since $CI$ is the perpendicular bisector of $DL$ we have $\angle A/2+\angle B=\angle IDC=\angle ILC=90+\angle A/3$ which gives $\angle B=60+\angle C/3$

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#5 h Mar 31, 2015, 8:07 AM • 2 Y

Y by Adventure10, Mango247

prove that
(sin b+c)^2 .sin b =(sin b - sin c)(sinb^2+sinc^2)

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#7 h Mar 16, 2018, 5:01 PM • 2 Y

Y by Adventure10, Mango247

I present a short sketch and leave the calculations to the reader: denote $AB=c\wedge BC=a\wedge CA=b\wedge x=AD$ . Then $AI=\frac{b+c}{a+b+c}\cdot x\wedge CD=\frac{ab}{b+c}$ . So $AI+CD=AC\iff x=b-\frac{a^2b}{(b+c)^2}$ . It's also possible to prove $AB=AD\iff c=b-\frac{a^2b}{(b+c)^2}$ .
Let's go to the merits.
If $AB=AD$ holds which is equivalent to $c=x$ , then $x=c=b-\frac{a^2b}{(b+c)^2}\implies AI+CD=AC$ .
If $AI+CD=AC$ holds, then $x=b-\frac{a^2b}{(b+c)^2}\iff \sqrt{cx}=\frac{\sqrt{(b+c-a)(a+b+c)}}{\frac{b+c}{\sqrt{bc}}}$ . The last part is an formula for bisector, so $\sqrt{cx}=AD=x\implies c=x\iff AB=AD$ QED

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#8 h Mar 16, 2018, 5:10 PM • 2 Y

Y by Adventure10, Mango247

Official solution.
Let $E$ be an interior point of segment $AC$ such that $AI=AE$ . Then $AI+CD=AC\iff CE=CD\iff \delta CDI \equiv \delta CEI \iff \angle AEI=\angle ADB\iff \delta AEI \sim \delta ADB \iff \angle ABC=\angle ADB\iff \angle B=\frac{1}{2}\angle A+\angle C\iff \angle B=60^{\circ}+\frac{1}{3}\angle C$

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#9 h Mar 9, 2019, 1:59 AM • 3 Y

Y by lolmanc123, Adventure10, Mango247

Construct $\omega$ with radius $CD$ at $C$ and let $\omega \cap AC=D'$ , then, $\angle AD'I=C+\frac{A}{2}$ and $\angle AID'=B$ , then,
$B=C+\frac{A}{2} \implies \frac{3B}{2}=\frac{180^{\circ}+C}{2} \implies B=60^{\circ}+\frac{1}{3}C$ Now showing the condition $AI=AD'$ is easy from $B=60^{\circ}+\frac{1}{3}C$ which can be done using trig bash!

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#10 h Aug 22, 2023, 6:36 PM

Y by

B=60°+C/3
3B-C=180°
A+B+C=180°
Equating these 2 equations,
A/2=B-C
In ∆ADC,
By sine rule,
sin(angle ADC)/AC=sin ( angle DAC)/CD
sin(B+A/2)/AC=sin( A/2)/CD=
{sin(B+A/2)- sin A/2}/ AC-CD
AC-CD/ AC={sin(B+A/2)- sin A/2}/ sin(B+A/2)
=2cos {(B+A)/2} sin (B/2) / sin (2B-C)
2B-C=180°-B
So,
AC-CD/ AC= 2sin (C/2) sin (B/2) / sin(180°-B)
=sin (C/2)/ cos(B/2)
In ∆AIC,
sin ( angle AIC)/ AC=sin ( angle ACI)/AI
AI/ AC=sin(C/2) / sin (B+A/2+C/2)
B+A/2+C/2=2B-C/2=(4B-C)/2=90°+(B/2)
So,
AI/AC=sin(C/2) / sin (90°+B/2)
=sin (C/2) / cos(B/2)
AC-CD / AC=AI/AC
AC-CD=AI
AI+CD=AC
@ Krishijivi

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