equal angles starting with a parallelogram ABCD with <DAC=90^o

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equal angles starting with a parallelogram ABCD with <DAC=90^o

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Source: Iranian Geometry Olympiad 2018 IGO Intermediate p5

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30632 posts

#1 h Sep 19, 2018, 8:34 PM • 3 Y

Y by Adventure10, Mango247, Rounak_iitr

Suppose that $ABCD$ is a parallelogram such that $\angle DAC = 90^o$ . Let $H$ be the foot of perpendicular from $A$ to $DC$ , also let $P$ be a point along the line $AC$ such that the line $PD$ is tangent to the circumcircle of the triangle $ABD$ . Prove that $\angle PBA = \angle DBH$ .

Proposed by Iman Maghsoudi

This post has been edited 2 times. Last edited by parmenides51, Sep 20, 2018, 9:23 AM
Reason: Proposed

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3042 posts

#2 h Sep 19, 2018, 8:42 PM • 4 Y

Y by math-o-enthu, Maths_Guy, mijail, Adventure10

anantmudgal09: Isogonal line lemma by constructing a point $Q$ which is the intersection of tangents at $B$ and $D$ to $(ABD)$ .

Me: 3- page trig/angle chase is enough for this (the solution I submitted

).

This post has been edited 1 time. Last edited by TheDarkPrince, Sep 20, 2018, 7:05 AM

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698 posts

#3 h Sep 19, 2018, 8:43 PM • 6 Y

Y by Pluto1708, coldheart361, zuss77, SenatorPauline, Aryan-23, Adventure10

Aren't $P$ and $H$ isogonal conjugates wrt $\triangle{DAB}$ ?

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1002 posts

Afo

#4 h Sep 9, 2019, 3:44 PM • 1 Y

Y by Adventure10

How do you this with trigonometry?

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Mahdi_Mashayekhi

694 posts

Mahdi_Mashayekhi

#6 h Dec 29, 2021, 8:36 AM

Y by

A solution with only angle chasing and similarity. the whole idea is to add circle PBD and some points.

Let AD and AB meet circle PBD at S and K. we will prove ∠PBD = ∠ABH.
Notice that instead of proving ∠PBD = ∠ABH we can prove SAP and BAH are similar.

step1 : ∠ADP = ∠CDB and ∠DAH = ∠PAK.
∠ADP = ∠ABD = ∠CDB
∠DAH = ∠ACD = ∠CAB = ∠PAK

step2 : PAK and ADH are similar.
∠PAK = ∠DAH and ∠AKP = ∠BDP = ∠HDA.

step3 : SAP and BAH are similar.
AP/AH = AK/AD = AS/AB and ∠SAP = 90 = ∠HAB

now we have ∠PBD = ∠PSA = ∠ABH so ∠PBA = ∠DBH is wanted.
we're Done.

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crazyeyemoody907

450 posts

crazyeyemoody907

#7 h Jun 9, 2023, 12:35 PM • 1 Y

Y by v4913

Solved with awesomehuman, CyclicISLscelesTrapezoid, CT17, razmath, v4913, on the 1st floor (mod 4) of a certain dorm.
$[asy] //23mophw11 (main) //setup; size(6cm); pen blu,grn,blu1,blu2,lightpurple; blu=RGB(102,153,255); grn=RGB(0,204,0); blu1=RGB(233,242,255); blu2=RGB(212,227,255); lightpurple=RGB(234,218,255);// blu1 lighter //defn pair A,B,C,D,M,H; A=(0,0); B=(4,-3); C=(4,0); D=A+C-B; M=C/2; H=foot(A,C,D); pair D1,H1; D1=2*A-D; H1=2*D-H; pair inverse(pair A, pair O,real r) { /*w = circle*/ real d=distance(A,O); return O+(A-O)*r*r/(d*d);} pair K=inverse(M,circumcenter(A,B,D),circumradius(A,B,D)); pair A1=inverse(A,K,distance(K,B)); //draw filldraw(A--B--C--D--cycle,blu1,blu); draw(A--C^^B--D^^K--A1,blu); draw(circumcircle(A,B,D),purple); draw(B--D1--A,blu+dashdotted); draw(B--K--D,magenta); //label void pt(string s,pair P,pair v,pen a){filldraw(circle(P,.07),a,linewidth(.3)); label(s,P,v);} pair points[] ={A,B,C,D,H,D1,A1,K}; //19 string labels[]={"$A$","$B$","$C$","$D$","$H$","$D'$","$A'$","$K$"}; //9 int dirs[]={170,-90,20,160,100,-90,-90,230}; pen colors[] ={blu,blu,blu,blu,blu, blu,purple,magenta}; for (int i=0; i< 8; ++i) { pt(labels[i], points[i], dir(dirs[i]), colors[i]); } [/asy]$
Define $K=\overline{DP}\cap\overline{AH}$ , so that we may use DDIT on $B$ and quadrilateral $AHDP$ . This gives us an involutive pairing
$(\overline{BC},\overline{BK};\overline{BP},\overline{BH};\overline{BA},\overline{BD})$ (this is NOT a cross ratio!)
Next, take the following reductions (i.e. ``equivalent-to-show'' 's):

As we want this involution to be a regular reflection in some line through $B$ , it's equivalent to show $\measuredangle KBA=\measuredangle DBC$ , the latter of which we may rewrite as $\measuredangle BDA$ ;
$\measuredangle BDA=\measuredangle KBA$ is equivalent to `` $\overline{KB}$ touches $(ABD)$ '';

It remains to prove that $\overline{AH}$ is a symmedian in $\triangle BAD$ . To finish, observe that $\angle DAH=\angle ACD=\angle CAB$ , so $\overline{AH}$ is isogonal to the median $\overline{AC}$ , completing the proof.

This post has been edited 6 times. Last edited by crazyeyemoody907, Jun 11, 2023, 2:09 AM

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#8 h Sep 16, 2023, 6:52 PM

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Let $E = AP \cap AH, AH \cap (ABD) = X$ . Then, by isogonals lemma it enough to prove that $BE$ and $BC$ are isogonal lines onto $BA, BD$ or since $\angle DBC = \angle ADB$ we need to prove that $EB$ is tangent to $(ADB)$ . Let's look: we need to prove that tangents to $(ADXB)$ at points $D,B$ are concur at $AX$ . So, we need to prove that $ADXB$ is garmonic quadrilateral or $\frac{AD}{AB}=^?\frac{DF}{BX}=^{\angle BAX = 90^{\circ}} sin \angle DAH = \frac{DH}{AD}$ or $AD^2=DH \cdot DC (DC=AB)$ which is follows since $\angle DHA = \angle DAC = 90^{\circ}$ $\blacksquare$ .

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