equal angles when <C = <A + 90^o

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equal angles when <C = <A + 90^o

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Source: IGO 2014 Junior 4

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30629 posts

#1 h Jul 22, 2018, 12:48 PM • 3 Y

Y by Adventure10, Mango247, Rounak_iitr

In a triangle ABC we have $\angle C = \angle A + 90^o$ . The point $D$ on the continuation of $BC$ is given such that $AC = AD$ . A point $E$ in the side of $BC$ in which $A$ doesn’t lie is chosen such that $\angle EBC = \angle A, \angle EDC = \frac{1}{2} \angle A$ . Prove that $\angle CED = \angle ABC$ .

by Morteza Saghafian

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Bikey

#2 h Jul 22, 2018, 7:49 PM • 3 Y

Y by Adventure10, Mango247, Rounak_iitr

I'm not sure from my solution ,but I think it's convincing .
Extend AB and let M the intersection of [BA) and the cCFircle (A,AC) MC is parallel to the besictor of BAC (Thales ) then CMA=A/2. And we have DMC =A . so
DMA=A+A/2=3A/2 (1) .

Now entend [BE) to Z . we have ZED=EDB+EBD =A+A/2 =3A/2 (2).
from 1 and 2 we conclude DEBM is cyclic
It remains to prove that E,C and M are collinear, but we have CMB =BDE and DBE= CMD ,so they must be collinear . and we are done .
Note : if my way is wrong please tell me .

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historypasser-by

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historypasser-by

#3 h Jul 23, 2018, 3:53 AM • 2 Y

Y by Adventure10, Mango247

use angle form of ceva
click to see more detail
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This post has been edited 4 times. Last edited by historypasser-by, Jul 23, 2018, 6:29 AM
Reason: spelling

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toanhocmuonmau123

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toanhocmuonmau123

#4 h May 1, 2020, 7:16 AM

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May you tell me the source of this problem? Thank you so much.

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toanhocmuonmau123

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toanhocmuonmau123

#5 h May 1, 2020, 7:24 AM

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toanhocmuonmau123 wrote:

May you tell me the source of this problem? Thank you so much.

Sorry, I have not read carefully.

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#6 h Apr 28, 2021, 4:28 PM

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My partial Solution

Attachments:

This post has been edited 2 times. Last edited by Enthurelx, Apr 28, 2021, 4:57 PM

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#7 h Sep 24, 2021, 4:55 AM

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EHIC isn't cyclic. Because by angle chasing AC is perpendicular to line BE. Then GCIH is cyclic

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4378 posts

#8 h Sep 24, 2021, 4:42 PM

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A good solution at my blog

Best regards,
sunken rock

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trying_to_solve_br

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trying_to_solve_br

#9 h Sep 24, 2021, 6:03 PM • 1 Y

Y by betongblander

Nice one, but pretty tricky, because what you need to do is take de perpendicular bisector of $CD$ and let it intersect $ED$ in $G$ and $BE$ in $H$ . Also, let $P=EC \cap AB$ . Notice that we only need to have $BEDP$ cyclic. Obviously $C$ is the orthocenter of $HAB$ , and thus $\angle CHE=A=\angle CGE=\angle GBC + \angle GCB$ . Thus $ECGH$ is cyclic and as $\angle CHA=90-2A$ (because $C$ is the orthocenter) this implies $\angle CED=\angle 90-2A$ and as $\angle CGE=90-3A/2$ , we have $\angle PCA=A/2$ which implies $EC$ parallel to the $A-$ bisector and thus $BEDP$ cyclic.

This sol is not fully mine; I just posted it here because the ones I've seen here were wrong/incomplete

This post has been edited 2 times. Last edited by trying_to_solve_br, Sep 24, 2021, 6:06 PM

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#10 h May 8, 2023, 2:24 PM

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trying_to_solve_br wrote:

$\angle CHA=90-2A$ (because $C$ is the orthocenter) this implies $\angle CED=\angle 90-2A$ .

You can end here. Also there's no need to prove $BEDP$ cyclic.

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#11 h Dec 5, 2023, 10:22 AM

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My solution is a trig bash:
Let angle CED=x
angle CDE=A/2
so, angle ECB=x+A/2
angle CBE=A
So, angle BEC=180°-{(3A/2)+x}
AC=AD, C=A+90°
B=90°-2A, angle CAD=2A
In ∆ABD,
sin 3A/ BD= sin(90°-2A)/AD
In ∆ABC,
sin A/ BC= sin(90°-2A)/AC
sin 3A/ BD= sin A/ BC
So, if we just evaluate it,
CD/ BC=2 cos 2A
In∆ECD,
EC=CD sin(A/2) / sin x
In∆BEC,
EC=BC sin A / sin{(3A/2)+x}
After evaluation,
2 cos 2A sin{(3A/2)+ x}= 2 sin x cos A/2
sin{(7A/2)+ x} - sin(x+A/2}=0
2cos(x+2A) sin(3A/2)=0
If sin( 3A/2)=0
A≠0
3A/2=180°
A=120°
C=210°( not possible)
cos(x+2A)= 0
x+2A=90°
x=90°-2A=B
x=angle CED
So, angle CED= B
@ Krishijivi

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#12 h Dec 8, 2023, 8:31 PM • 2 Y

Y by starchan, Rounak_iitr

Let $K$ be on the extension of $BA$ such that $AC=AD=AK$ . (This implies $\angle AKC = \angle ACK = \frac{A}{2}$ .) Let the tangent to $(ABC)$ at $B$ intersect $KC$ at $E'$ . Then
$\angle E'KD = \angle CKD = \frac{\angle CAD}{2} = A = \angle E'BD \implies BE'DK$ cyclic.

$\angle E'DC = \angle BDE' = \angle BKE' = \angle AKC = \frac{A}{2} \implies E'=E$ .

Therefore
$\angle CED = \angle KED = \angle KBD = \angle ABC$ as desired.

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#13 h Dec 17, 2024, 5:49 AM

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