Show that the sum of the angles is equal to 180

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Show that the sum of the angles is equal to 180

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Source: Turkey JBMO TST 2014 P1

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#1 h May 30, 2014, 7:21 PM • 2 Y

Y by FaThEr-SqUiRrEl, Adventure10

In a triangle $ABC$ , the external bisector of $\angle BAC$ intersects the ray $BC$ at $D$ . The feet of the perpendiculars from $B$ and $C$ to line $AD$
are $E$ and $F$ , respectively and the foot of the perpendicular from $D$ to $AC$ is $G$ . Show that $\angle DGE + \angle DGF = 180^{\circ}$ .

This post has been edited 1 time. Last edited by Amir Hossein, Jun 18, 2018, 8:59 AM

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#2 h May 31, 2014, 2:38 AM • 3 Y

Y by FaThEr-SqUiRrEl, Adventure10, Mango247

Prolongamos $DG$ , mas alla de $G$ , hasta el punto $X$ .
Si $\angle DGE + \angle DGF=180$ entonces $\angle DGE=\angle FGX$ (*)

$FC\parallel EB$ entonces $\angle FCD=\angle EBD= \alpha$
$DCFG$ es ciclico, luego $\angle FGX=\alpha$ (1)

$P=EB\cap AC$ , entonces $AB=AP$ y $DB=DP$ , luego $\angle DPE=\alpha$ .
$DEGP$ es ciclico,
como $\angle PDE=90-\alpha$ entonces $\angle PGE=90+\alpha$ .
Luego $\angle DGE=(90+\alpha)-90=\alpha$ (2)

(1)=(2)=(*)

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Mikasa

#3 h May 31, 2014, 6:29 AM • 4 Y

Y by IstekOlympiadTeam, FaThEr-SqUiRrEl, Adventure10, Mango247

Let the internal bisector of $\angle BAC$ intersect $BC$ at $T$ . Let the rays $DG$ and $EB$ intersect at $S$ .

Now $\angle CAT=\dfrac{\angle A}{2}, \angle CAF=\dfrac{180^{\circ}-\angle A}{2}=90^{\circ}-\dfrac{\angle A}{2}$ . So, $\angle TAF=90^{\circ}=\angle TAE$ . Since $\angle CAT=\angle BAT$ , we have that $\angle EAB=\angle FAC$ .

Now since $\angle AES+\angle AGS=180^{\circ}$ we have that $E,A,G,S$ are concyclic. So,
$\angle BSD=\angle ESG=180^{\circ}-\angle EAG=\angle FAC=\angle EAB=180^{\circ}-\angle BAD$ .
So $B,A,D,S$ are cyclic as well.

Now,
$\EGD=180^{\circ}-\angle EGS=180^{\circ}-\angle EAS=\angle SAD=\angle SBD=180^{\circ}-EBD=180^{\circ}-\angle FCD$.
But $\angle CFD+\angle CGD=180^{\circ}$ means that $C,F,D,G$ are concyclic. So, $\angle FCD=\angle FGD$ .

Thus $\angle EGD=180^{\circ}-\angle FGD$ , as desired.

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MrRTI

#4 h May 31, 2014, 7:16 AM • 3 Y

Y by FaThEr-SqUiRrEl, Adventure10, Mango247

Let the internal angle bisector of $\angle{BAC}$ meets $BC$ at $D'$ then the ratio $(B, C ; H, D)$ is harmonic and because $F, E, A, D$ are the perpendicular projections from $B, C, D', D$ respectively, then the ratio $(F, E; A, D)$ is also harmonic.
Since $\angle{AGD} = 90^{\circ}$ and $(F, E; A, D) = -1$ then $GA$ bisects $\angle{EGF}$ .
Notice that $\angle{CED} + \angle{CGD} = 90^{\circ} + 90^{\circ} =180^{\circ}$ , we have $CGDE$ is cyclic, hence if $GF$ meets $BD$ at $P$ , we get $\angle{PGA} = \angle{CGE} = \angle{PDA}$ , so $ADGP$ is cyclic hence $\angle{APD} = 90^{\circ}$ . Thus, $APBF$ is cyclic, so $\angle{DGE} + \angle{DGF} = \angle{DCE} + \angle{PAF} = \angle{DBF} + \angle{PAF} = 180^{\circ}$ .

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bcp123

#5 h May 31, 2014, 1:06 PM • 3 Y

Y by FaThEr-SqUiRrEl, Adventure10, Mango247

Nothing complicated is required. $EAG$ is similar to $BAD$ then $\angle DGE=90+D$ and $\angle DGF=90-D$ .

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#6 h Jun 2, 2014, 10:59 PM • 2 Y

Y by FaThEr-SqUiRrEl, Adventure10

Another solution:
Let X be the projection of A on to BC.Now,we have two cyclic quadritedrals,GDAX and BXAE.Now,from cyclic quadritedral GDFC we have <DGF=<DCF=90-(<C-<B)/2.Also,<EXC=90+GXD+<EXA=90+<XAD+<XDA=180 => G,X and E are collinear => <EGA=<XGA=<XDA=(<C-<B)/2,and now the conclusion easily follows...

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#7 h Aug 19, 2016, 1:52 AM • 2 Y

Y by FaThEr-SqUiRrEl, Adventure10

here is one line solution : DI pedencular AB then DGE=DIE=DBE=DCF=180-DGF ( symetric , AD bisector)

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#8 h Mar 15, 2019, 5:34 PM • 3 Y

Y by FaThEr-SqUiRrEl, Adventure10, Mango247

Purely Projective!
Let $X$ be the intersection of $A-$ angle bisector with $BC$ , then,
$-1=(D,X;B,C) \overset{\infty_{AX}}{=} (D,A;E,F) \implies \angle EGA=\angle FGA \implies \angle DGF+\angle DGE=180^{\circ}$

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#9 h Jul 21, 2020, 7:33 PM • 1 Y

Y by FaThEr-SqUiRrEl

Solution without projective geometry (Same logic though :-D

:-D

)

First step is the same. Let the internal bisector of $\angle A$ intersects $BC$ at $X$ . I think this is the most important step.
So here is motivation
Furthermore we know that from similarity ( $EB // AX // FC$ ) and properties of bisectors
$\frac{EA}{AF} = \frac{BX}{XC} = \frac{DB}{DC} = \frac{DE}{DF}$ We can easily see that (from the properties of bisectors, again) $GA$ is the bisector of $\angle EGF$ . WLOG $\angle EGA = \alpha$ . Then we have that $\angle DGF = 90+\alpha \; \; \angle DGE = 90- \alpha \Rightarrow \angle DGF + \angle DGE= 180$ .

This post has been edited 1 time. Last edited by electrovector, Jul 21, 2020, 7:34 PM

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hakN

#10 h Apr 8, 2021, 9:46 AM • 1 Y

Y by FaThEr-SqUiRrEl

Mine is very brute force solution.

Note that it is equivalent to show $\angle DGE = \angle EBD$ .
By the ratio lemma, we get $\frac{DE}{DF} = \frac{GE\cdot \sin{\angle DGE}}{GF\cdot \sin{(\angle C + \angle{\frac{A}{2}})}} = \frac{DB}{DC} = \frac{AB}{AC} = \frac{\sin{\angle C}}{\sin{\angle B}}$ .
We also know $\frac{GE}{GF} = \frac{\sin{\angle C}}{\sin{\angle GED}}$
So, $\frac{\sin{\angle DGE}}{\sin{\angle GED}} = \frac{\sin{(\angle C + \angle{\frac{A}{2}})}}{\sin{\angle B}}$
But we also know that $\angle DGE + \angle GED = 180 - \angle{\frac{A}{2}} = (\angle C + \angle{\frac{A}{2}}) + \angle B$ .
So, by a well known lemma, we get that $\angle DGE = \angle C + \angle{\frac{A}{2}} = \angle EBD$ . Done!

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