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 ).

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

How do you this with trigonometry?

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.

Solved with awesomehuman, CyclicISLscelesTrapezoid, CT17, razmath, v4913, on the 1st floor (mod 4) of a certain dorm. [asy][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][/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.

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 $\angle DHA = \angle DAC = 90^{\circ}$ $\blacksquare$ .