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#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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#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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#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.

. i award myself the 1st position in foolishness competition 2012

This post has been edited 5 times. Last edited by mathbuzz, Aug 18, 2012, 6:27 PM

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#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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#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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#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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#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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#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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#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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#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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