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equal segments on radiuses
danepale   8
N 41 minutes ago by zuat.e
Source: Croatia TST 2016
Let $ABC$ be an acute triangle with circumcenter $O$. Points $E$ and $F$ are chosen on segments $OB$ and $OC$ such that $BE = OF$. If $M$ is the midpoint of the arc $EOA$ and $N$ is the midpoint of the arc $AOF$, prove that $\sphericalangle ENO + \sphericalangle OMF = 2 \sphericalangle BAC$.
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danepale
Apr 25, 2016
zuat.e
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B.Math - geometry
mynamearzo   9
N Apr 25, 2025 by Apple_maths60
In a triangle $ABC$ , $D$ is a point on $BC$ such that $AD$ is the internal bisector of $\angle A$ . Now Suppose $\angle B$=$2\angle C$ and $CD=AB$ . Prove that $\angle A=72^0$.
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mynamearzo
Jun 17, 2012
Apple_maths60
Apr 25, 2025
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mynamearzo
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mynamearzo
#1 h  Jun 17, 2012, 11:18 AM • 2 Y
Y by Adventure10, Mango247
In a triangle $ABC$ , $D$ is a point on $BC$ such that $AD$ is the internal bisector of $\angle A$ . Now Suppose $\angle B$=$2\angle C$ and $CD=AB$ . Prove that $\angle A=72^0$.
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j666
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j666
#2 h  Jun 17, 2012, 11:27 AM • 2 Y
Y by Adventure10, SatisfiedMagma
Suppose, $ BE $ is the angle-bisector of $ \angle ABC $. From the given condition we get that, $ \Delta CDE\equiv \Delta ABE $. So $ DE=AE $. So $ \angle EDA=\angle DAE\implies DE\parallel AB\implies AC=BC $. So $ \angle BAC=72 $. So done.
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mathbuzz
803 posts
mathbuzz
#3 h  Aug 14, 2012, 2:40 PM • 2 Y
Y by Adventure10, Mango247
let $\angle C=\theta$. then $B=2\theta$ and $A=\pi-3\theta$
$\angle BAD=\frac{\pi-3\theta}{2}$
now, as $BD/DC=AB/AC$, and $DC=AB$ , we have , $ab=c^2+bc$ [symbols have their usual meanings]
now $B=2C$ , so , $b^2=c(c+b)$. so , it leads to $a=b$. so , $\pi-3\theta=2\theta$ and so , $A=$$72^\circ$

edit-- really sorry , did completely perfect in the copy, but made a typo while posting.
i just got $ab=bc+c^2$ in copy and missed $bc$ while typing.:oops: . i award myself the 1st position in foolishness competition 2012 :first: :rotfl:
This post has been edited 5 times. Last edited by mathbuzz, Aug 18, 2012, 6:27 PM
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sunken rock
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sunken rock
#4 h  Aug 15, 2012, 5:07 PM • 1 Y
Y by Adventure10
@mathbuzz:
You do not have $ab=c^2$, but $b\cdot BD=c^2$; keeping in mind that $BD=\frac{ac}{b+c}\implies\frac{abc}{b+c}=c^2\iff ab=c(b+c)$. This one, combined with $b^2=c(b+c)$leads, indeed, to $a=b$, a.s.o.

Best regards,
sunken rock
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partho8897
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partho8897
#5 h  Feb 14, 2015, 8:38 AM • 3 Y
Y by prerna123, Adventure10, Mango247
mathbuzz how is BD/DC=AB/AC?
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Submathematics
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Submathematics
#6 h  Feb 14, 2015, 10:11 AM • 2 Y
Y by Adventure10, Mango247
$\frac{BD}{DC}=\frac{AB}{AC}$ is the internal angle BISECTOR THEOREM.
See here http://en.wikipedia.org/wiki/Angle_bisector_theorem
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Abhinandan18
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Abhinandan18
#7 h  Dec 31, 2015, 5:24 AM • 2 Y
Y by Adventure10, Mango247
Well this is how I did it...
Applying sine rule to $\Delta ADB$ and $\Delta ADC$ we have, $\frac{AD}{\sin B}=\frac{AB}{\sin \angle ADB}$ and $\frac{AD}{\sin C}=\frac{CD}{\sin \frac{A}{2}}$
$\therefore \frac{\sin B}{\sin ( C+\frac{A}{2})}=\frac{\sin C}{\sin \frac{A}{2}}$ and then plugging in the given criteria we have $\cot C=\cot \frac{A}{2}$ and thus $A=2C$ (as they are angles of a triangle). Hence, $A=B=72^\circ$ and $C=36^\circ$
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Eduline
862 posts
Eduline
#8 h  Apr 29, 2019, 7:24 AM • 1 Y
Y by Adventure10
Another Solution: Let $\angle ACB=x$, which implies that $\angle ABC=2x$. Now, by sine rule on $\Delta ABC$ we have $$\frac{AB}{AC}=\frac{\sin x}{\sin 2x}\implies \frac{DC}{AC}=\frac{\sin x}{\sin 2x}.$$
Observe that $\angle DAC=90^0-\frac{3x}{2}$ and $\angle ADC=90^0+\frac{x}{2}$. Now again by sine rule on $\Delta ADC$, we have $$\frac{DC}{AC}=\frac{\cos\frac{3x}{2}}{\cos\frac{x}{2}}.$$
Therefore we have $\sin x.\cos\frac{x}{2}-\sin 2x.\cos\frac{3x}{2}=0$

$\implies \sin x.\cos\frac{x}{2}-2\sin x .\cos x. \cos\frac{3x}{2}=0$

$\implies \cos\frac{x}{2}-2 \cos x. \cos\frac{3x}{2}=0$ ($\sin x\neq 0$, since $x\neq 0$).

$\implies \cos\frac{5x}{2}=0$.

Therefore $x=\frac{\pi}{5}$.

Therefore $\angle BAC=180^0-3x=72^0.$
This post has been edited 3 times. Last edited by Eduline, Apr 29, 2019, 7:30 AM
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SatisfiedMagma
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SatisfiedMagma
#11 h  Dec 10, 2023, 8:15 AM
Y by
Okay, two finishes. Synthetic one was smooth though! Even the trigonometric solution is pretty cool as well.

Solution: Let $I$ be the incenter of $\triangle ABC$. Observe that $\triangle BID$ is isosceles with $\overline{BI} = \overline{BD}$ since
\[\angle BDI = \angle DAC + \angle DCA = \angle IAB + \angle IBA = \angle BID.\]Call $E \coloneqq BI \cap AC$. From here, we have two finishes.

Synthetic Solution It is easy to see that $\overline{EB} = \overline{EC}$ since $\angle EBC = \angle ECB$. By SAS criteria, we have that $\triangle EBA \cong \triangle ECD$. Now observe
\[\angle DEC = \angle BEA = \angle EBD + \angle ECB = 2\angle EBC = \angle ABC\]which means $DEAB$ is cyclic. This finally means $\angle DAE = \angle DBE$ implying $\triangle CBA$ is isosceles with $\overline{CB} = \overline{CA}$. If you call $\angle BCA = \theta$, then $\angle CBA = \angle CAB = 2\theta$. Finally, by angle sum in $\triangle ABC$, we have
\[\theta + 2\theta + 2\theta = 180^\circ \iff \angle A = 2\theta = 72^\circ.\]and now we're done. $\blacksquare$

Trigonometric Finish From angle bisector theorem,
\[\frac{AB}{AC} = \frac{BD}{DC} = \frac{BI}{BA}.\]Let $x = \angle ECB$, $y = \angle DAC = \angle DAB = y$, then applying Sine Rule in $\triangle ABC$ and $\triangle BIA$ in light of above equation, we get
\[\frac{\sin(x)}{\sin(2x)} = \frac{\sin(y)}{\sin(x+y)}.\]Expand with double angle and compound angle formulae to get
\begin{align*} \frac{\sin(x)}{2\sin(x)\cos(x)} & = \frac{\sin(y)}{\sin(x)\cos(y) + \sin(y)\cos(x)} \\ \iff 2\cos(x)\sin(y) & = \sin(x)\cos(y) + \sin(y)\cos(x) \\ \iff \tan(x) & = \tan(y) \end{align*}Since $x,y$ are angle of triangle, we can conclude that $x = y$. An analogous calculation as above would give $\angle BAC = 72^\circ$. This finishes the solution. $\blacksquare$
This post has been edited 2 times. Last edited by SatisfiedMagma, Jan 13, 2024, 3:21 PM
Reason: removed wrong diag
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Apple_maths60
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Apple_maths60
#15 h  Apr 25, 2025, 1:34 PM
Y by
By Angle Bisector Theorem

$\frac{AB}{AC}=\frac{BD}{DC}$

Applying Sine rule we get $\frac{AB} {\sin D}=\frac{AD}{\sin B}$ from $ \Delta ABD$

So, now $AD = \frac{AB \sin B}{\sin D}$ Similarly, $\frac{DC}{\sin \frac{A}{2}} =\frac{AD}{\sin C}$ from $\Delta ACD$

Plugging the value of $AD$ in the above equation we get, $\sin C. \sin D = \sin B.\sin\frac{A}{2}$

$\implies\sin C \sin (\frac{A}{2} +C) = \sin 2C. \sin \frac{A}{2}$

$\implies \sin C. \cos\frac{A}{2} + \cos C. \sin\frac{A}{2} =2\cos C. \sin\frac{A} {2}$

$\implies \sin C. \cos\frac{A}{2} = \cos C. \sin\frac{A}{2}$

$\implies \tan C=\tan \frac{A}{2}$

Thus $C=\frac{A}{2} \implies 2C = A$ So, $ 5C =\pi \implies C =36^\circ$

Thus we get $A=72^\circ$
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