I will post the official solution:

Nice one! Let $F$ be the point on $BC$ such that $CF = CB = CD$. This actually makes $CFD$ and equilateral triangle and so now a bit of angle chase to show that $\angle FAC = \angle FCA = 35^\circ$ and so $FA = FC$ and now some more angle chase to get that $\angle DAC = 30^\circ$. Then, just use law of sines to get that $CE = BD$

Proceeds to Trig Bash

Redxated

Cheers @above

@anantmudgal09, Is there a way by which we can draw the figure only with a pencil, straightedge, a compass? This is because I don't know how to draw an angle of $25^{\circ}$

yeah @2above I bashed too...

I was simplifying my trig bash somehow by Ptolemy and random similarities...I saw the examiner taking other's paper...luckily I was in the very last seat and somehow fakesolved xd

N1RAV wrote: @anantmudgal09, Is there a way by which we can draw the figure only with a pencil, straightedge, a compass? This is because I don't know how to draw an angle of $25^{\circ}$ There is a method to approx a 20 degree angle (it gives 19.1 degree which you can use) Search google/YouTube for it

Let the perpendicular bisector of $BD$ intersect $AB$ at $F$. We get 1. $\angle FCD = \frac{1}{2} \angle BCD = 50^{\circ}$ and since $\angle ACD = 25^{\circ}$, $CA$ bisects $\angle FCD$. 2. $\angle BFC = 180^{\circ} - \angle FBC - \angle FCB = 60^{\circ}$. Due to symmetry, $\angle CFD = \angle BFC = 60^{\circ}$. Therefore, $AF$ must be the external angle bisector of $\angle DFC$. Now, observe $\triangle CFD$. Notice that $A$ is the intersection of internal angle bisector of $\angle FCD$ and external angle bisector $\angle CFD$ and so $A$ must be $C$-excenter of $\triangle CFD$. So, $AD$ must be the external angle bisector of $\angle CDF$ and hence $\angle FDA = 90^{\circ} - \frac{1}{2} \angle FDC = 90^{\circ} - \frac{1}{2} \angle FBC = 55^{\circ}$. Therefore, $\angle CDA = \angle CDF + \angle FDA = 125^{\circ}$. Now, with a little more angle chase, we can find that $\angle CED = \angle CAD = 30^{\circ}$ and $\angle EDC = 50^{\circ}$. By law of sine on $\triangle ECD$, we get $\frac{EC}{CD} = \frac{\sin 50^{\circ}}{\sin 30^{\circ}} = 2\sin 50^{\circ}$. But from isosceles triangle $\triangle BCD$, it is clear that $\frac{BD}{2 BC} = \sin 50^{\circ}$. Since $BC = CD$, we conclude that $BD = EC$.

Pls somebody send figure fir the prblem

Piyushkhattar wrote: Pls somebody send figure fir the prblem

My trig bash

Thnx L567

Piyushkhattar wrote: Pls somebody send figure fir the prblem https://youtu.be/yN-Az2OMW_s

anantmudgal09 wrote: In a convex quadrilateral $ABCD$, $\angle ABD=30^\circ$, $\angle BCA=75^\circ$, $\angle ACD=25^\circ$ and $CD=CB$. Extend $CB$ to meet the circumcircle of triangle $DAC$ at $E$. Prove that $CE=BD$. Proposed by BJ Venkatachala Really Amazing geo...I am in love with this now.. Here's My Solution... Let $BD\cap AC=H$ Now Take Point $J$ such that $\angle CAJ=5^\circ , DBJ=25^\circ$ Claim 1-: $\angle AJH=10^\circ, \angle BJH=75^\circ$ Proof -: Construction $O$ As circumcentre of $\triangle AJB$ Now take point $H^*\in BH$ Such that $JH^*OB$ is Concyclic Now observe as $BH^*$ bisects $\angle OBJ$ So $JH^*=OH^*, \angle H^*JO=25^\circ \angle H^*HJ=25^\circ$ Now Contruct $K$ inside $\triangle AH^*O$ such that $\triangle KH^*O$ Is equilateral Then observe as $AO=OJ$ and $\angle AOH=110^\circ$ So $\angle AOK=\angle H^*OJ=25^\circ$ With $KO=OH^*\implies \triangle AOH \cong \triangle OH^*J$ which give us $KA=JH^*=KH^*=KO$ Hence $\angle JAH^*=5^\circ$ But we have $\angle JAH=5^\circ$ Hence $\boxed{H^*\equiv H}$. Hence our claim is proved. Claim 2-: $J\in DC$ Proof -: As we have $\angle HJB=75^\circ, \angle HBJ=25^\circ$ So $JHBC$ Is Concyclic. Hence $J\in DC$ Claim 3-: $ADBJ$ is Concyclic Proof-: As $J\in DC$ So $\angle AJC=65\circ+75^\circ+10^\circ=150^\circ$ So $\angle AJD=30^\circ=\angle ABD$ Hence $ADBJ$ is Concyclic. $\implies \angle ADB=\angle AJB=85^\circ, \angle DAH=30^\circ$ Claim 4-: $CE=BD$ First observe as $ADCE$ is Concyclic So $\angle DEC=30^\circ$ Now Let $L$ Be the Circumcentre of $\triangle DCE$ Then we have $LD=LC=LE=DC=BC$ Now take another point $N\in EC$ such that $LE=BC=EA\implies \angle LNE=\angle NLE=20^\circ$ Hence $BN=CE$ And Finally observe $\triangle DBN$ is isosceles So $\boxed{BD=BN=CE}$ $\blacksquare$ Huh!

Probably my favorite question on today's INMO. I'm personally all for reviving this sort of old geo - Completely free from configurations/few points with unique conditions and a surprising statement. Also completely angle chase. An outline: Construct equilateral triangle $COD$ such that $\angle BCO = 100^\circ - 60^\circ = 40^\circ$. Define $A'$ on circle centred at $O$ through $C$ such that $\angle A'CD = 25^\circ$. The key claim is that $A = A'$. By angle chasing, one can note that $\angle A'OC = 180^\circ - 2\angle A'CO = 180^\circ - 70^\circ = 110^\circ$ Also observe that $\angle BOC = \angle OBC = 90^\circ - \frac{BCO}2 = 70^\circ$. Thus, $BOA'$ are collinear. Now, note that $\angle A'BC = \angle OBC = \frac{180^\circ-\angle OCB}{2} = \angle 70^\circ = \angle ABC$ which implies that $A' = A$ To finish, observe that $EOC \cong BCD$

...........

a solution given in Romantics of Geometry (fb group) by Apostolis Manoloudis, a greek maestro in angle-related problems (always synthetically) enjoy

Let $X$ be such that $BDX$ is equilateral.Then $BA\perp XD$ means $XAD$ is isosceles. Also since $CA$ bisects $\angle{XCD},XACD$ is cyclic. Also easily seen $XE\parallel CD$. Hence $CE=XD=BD$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10.0cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.4) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 3, xmax = 13, ymin =3, ymax = 12; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215686274509803,0); pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); draw((5.93582222752409,11.035252053883907)--(7,5)--(11.69459271066772,8.939231012048833)--cycle, linewidth(1) + zzttqq); draw(arc((11,5),0.545454545454545,80,105)--(11,5)--cycle, linewidth(1) + blue); draw(arc((11,5),0.545454545454545,105,130)--(11,5)--cycle, linewidth(1) + blue); draw(arc((7,5),0.545454545454545,40,70)--(7,5)--cycle, linewidth(1) + qqwuqq); draw((8.683292144015187,9.624807156027094)--(9.045726520954464,9.49289183094638)--(9.177641846035177,9.855326207885657)--(8.8152074690959,9.98724153296637)--cycle, linewidth(1) + qqwuqq); Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 1, Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 1, Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw((7,5)--(11.69459271066772,8.939231012048833), linewidth(1)); draw((11.69459271066772,8.939231012048833)--(11,5), linewidth(1)); draw((9.30390280082676,11.329920921889787)--(11,5), linewidth(1)); draw((9.30390280082676,11.329920921889787)--(11.69459271066772,8.939231012048833), linewidth(1)); draw((9.30390280082676,11.329920921889787)--(7,5), linewidth(1)); draw(circle((7.935822227524095,7.571150438746157), 4), linewidth(1)); draw((7,5)--(11,5), linewidth(1)); draw((4.87164445504819,5)--(7,5), linewidth(1)); draw((5.93582222752409,11.035252053883907)--(7,5), linewidth(1)); draw((5.93582222752409,11.035252053883907)--(11.69459271066772,8.939231012048833), linewidth(1)); draw((5.93582222752409,11.035252053883907)--(4.87164445504819,5), linewidth(1)); draw((5.93582222752409,11.035252053883907)--(7,5), linewidth(1) + zzttqq); draw((7,5)--(11.69459271066772,8.939231012048833), linewidth(1) + zzttqq); draw((11.69459271066772,8.939231012048833)--(5.93582222752409,11.035252053883907), linewidth(1) + zzttqq); draw((5.93582222752409,11.035252053883907)--(11,5), linewidth(1)); draw((5.93582222752409,11.035252053883907)--(9.30390280082676,11.329920921889787), linewidth(1)); /* dots and labels */ dot((7,5),dotstyle); label("$B$", (6.689090909090905,4.572727272727276), NE * labelscalefactor); dot((11,5),dotstyle); label("$C$", (11.234545454545447,4.681818181818185), NE * labelscalefactor); dot((11.69459271066772,8.939231012048833),dotstyle); label("$D$", (11.761818181818175,9.118181818181824), NE * labelscalefactor); dot((9.30390280082676,11.329920921889787),linewidth(4pt) + dotstyle); label("$A$", (9.38,11.48181818181819), NE * labelscalefactor); dot((4.87164445504819,5),linewidth(4pt) + dotstyle); label("$E$", (4.3618181818181805,4.645454545454549), NE * labelscalefactor); dot((5.93582222752409,11.035252053883907),linewidth(1pt) + dotstyle); label("$X$", (5.689090909090906,11.227272727272736), NE * labelscalefactor); label("$25^{\circ}$", (10.798181818181812,5.863636363636368), NE * labelscalefactor,blue); label("$25^{\circ}$", (10.343636363636357,5.718181818181822), NE * labelscalefactor,blue); label("$30^{\circ}$", (7.343636363636359,5.572727272727277), NE * labelscalefactor,qqwuqq); dot((8.8152074690959,9.98724153296637),linewidth(1pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Edit: Lol. solution of almost same size of the problem

@anantmudgal09 @anyone I initially chased angles and wrote some of them mentioned above , but then , I dived to trigonometry and simplified it to a good extent and in last it was just left to prove that the small expression consisting of some weird trigo ratios is equal to 1 ,Can I hop partial marks for it(how much ) , bcoz I read somewhere that no partial marks are given for partially solved trigo approach in INMO

Call $\angle ADC = \theta $ We are going to find out $\theta$ via trigonometry. Let $CB=CD=x$ We obtain $BD= 2CBsin(50^{\circ})$ (Cosine Rule) By Sine Rule in $\Delta ADB$ $AD= \frac{xsin(50^\circ)}{sin(35^{\circ}+ \theta)} $ Sine Rule (again...) in $\Delta ADC$ and combining above we get $sin(35+\theta)= 2cos(25^\circ)sin(\theta)\dots 1$ $\angle ADB= 115^{\circ} - \theta \implies \theta \leq 115^{\circ}\dots 2$ Also from 1 we have $sin^2(\theta)= \frac{sin^2(35^{\circ})}{(2sin(65^\circ)-cos(35^\circ))^2 +sin^2(35^\circ)}= \frac{1}{4} \implies \theta= 30^{\circ}$ Now define $B'= DB \cap (DAC) $ Now, in $\Delta CB'E$ we find that $\angle CB'E=100^\circ +\theta$ This gives $CE= \frac{x sin(10^{\circ}+\theta)}{sin(\theta)}= 2xsin(50^\circ) = BD$ and we are done!

Let $F$ onto $AC$ so that $\angle AFD=30^\circ$ and $O$ reflection of $D$ across $AC$; $\triangle CDO$ is equilateral. Clearly $BCDO$ is a kite, so $BO=DO=OF$, consequently, as $\angle BOC=\angle OBC=65^\circ$, we get $\angle BOF=70^\circ$, i.e. $O$ is the circumcenter of $\triangle ABF$. Additionally, as $\angle AFD=\angle ABD=30^\circ, ABFD$ is cyclic. From $\angle ABF=\angle ABO+\angle OBF=60^\circ$, we infer $\angle DAF=\angle DBF=30^\circ$. Therefore $\angle CED=\angle CAD=30^\circ$. If $M$ is midpoint of $BD$ and $N$ projection of $C$ onto $DE$, then $\triangle CDM\cong\triangle DCN$, hence $DM=CN$, but $BD=2DM=2CN=CE$, done. Best regards, sunken rock

choose $O$ on $AB$ such that $\triangle COD$ is equilateral so $\angle DOA=50$ so $O$ is the circumcenter of $\triangle CDA$ $\angle CEO =40=\angle CDB$ so $\triangle OCE =\triangle BCD$ so $EC = DB$ and we win

Here is a master method to solve such problems. The idea is to inscribe the configuration in a regular n-gon. We consider $36$-gon in this case with vertices $X_1,\cdots X_{36}$. Let $C=X_1$ and $D=X_9$. Then we observe that $B=X_{29}$, $X_{15}$ lies on $BA$ and $X_{14}$ lies on $CA$. Wishfully thinking, we want to prove $\angle{ADB}=85$, which is equivalent to $X_{12}$ lying on $DA$. Hence the problem is equivalent to $$\text{In a 36-gon with vertices }X_1,\cdots X_{36}\text{ we have } X_1X_{14},X_9X_{12},X_{29}X_{15} \text{ are concurrent}$$Which is straightforward by complex numbers. You might need the fact that 36th cyclotomic polynomial is $x^{12}-x^6+1$.

kapilpavase wrote: Here is a master method to solve such problems. The idea is to inscribe the configuration in a regular n-gon. We consider $36$-gon in this case with vertices $X_1,\cdots X_{36}$. Let $C=X_1$ and $D=X_9$. Then we observe that $B=X_{29}$, $X_{15}$ lies on $BA$ and $X_{14}$ lies on $CA$. Wishfully thinking, we want to prove $\angle{ADB}=85$, which is equivalent to $X_{12}$ lying on $DA$. Hence the problem is equivalent to $$\text{In a 36-gon with vertices }X_1,\cdots X_{36}\text{ we have } X_1X_{14},X_9X_{12},X_{29}X_{15} \text{ are concurrent}$$Which is straightforward by complex numbers. You might need the fact that 36th cyclotomic polynomial is $x^{12}-x^6+1$. Looking Amazing !! From where we can learn about this technique ??

Problem 1: A well-known problem Let $BCD$ be a triangle with $BC=CD$ and $\angle BCD = 100^\circ$. Let $E$ be a point on the ray $[CB$ but outside of segment $[BC]$ such that $\angle CED = 30^\circ$. Show that $BD = EC$.

Back to the original problem. To solve the problem, we will show $CAD = 30^\circ$. Since $EADC$ is cyclic, $\angle CED = \angle CAD = 30^\circ$. So this will take us to the above well-known problem. In fact it is a more general problem.. Problem 2: Let $ABCD$ be a quadrilateral such that $\angle ABD = 30^\circ$, $\angle BCA = 90^\circ - 3\alpha$, $\angle ACD = 30^\circ - \alpha$ and $CD = CB$. Show that $\angle CAD = 30^\circ$. For $\alpha = 5^\circ$, it is the same quadrilateral in the original problem. This type of problem can be easily converted to P inside ABC - Trigonometric Ceva Models. In fact, it belongs to the family - Model 1. However, I will give a proof to the general quadrilateral version - Problem 2.

Trigonometry- Let $a = BC = CD$ and $\angle DAC = \theta$. Claim 1: $ \theta = 30^{\circ}$ Proof: Consider $\triangle BDC$. It is easy to see through cosine rule that $BD = 2a\cos(100^{\circ})$ (1). Consider $\triangle ABD$. Through angle chasing we see that $\angle ABD = 115^{\circ} - \theta$. Join $A$ with $E$ to get a cyclic quadrilateral $ADCE$. Since sum opposite sides of a cyclic quadrilateral sum up to $180^{\circ}$, we have that, $\angle AEC = 25^{\circ} + \theta$. Now, consider $\triangle ABE$. Through angle chasing we see that $\angle BAE = 45^{\circ} = \theta$. Now, consider $\triangle ACB$. Using sine rule we see that $AC = a\frac{\sin 70^{\circ}}{\sin 35^{\circ}} \Rightarrow AC = 2a \sin 55^{\circ}$. Now, consider $\triangle ADC$. Using sine rule we see that, $\frac{a}{\sin \theta} = \frac{AC}{\sin (155^{\circ} - \theta)} \Rightarrow $ $\frac{a}{\sin \theta} = \frac{2a \sin 55^{\circ}}{\sin (25^{\circ} + \theta)} \Rightarrow \sin (25^{\circ} + \theta) = 2 \sin 55^{\circ} \sin \theta \Rightarrow \cos (65^{\circ} - \theta) = 2 \sin 55^{\circ} \sin \theta \Rightarrow \cos (65^{\circ} - \theta) = \cos (55^{\circ} - \theta) - \cos (55^{\circ} + \theta) \Rightarrow \cos (65^{\circ} - \theta) + \cos (55^{\circ} + \theta) = \cos (55^{\circ} - \theta)$ Using addition to product formula of cosine we have, $\cos (5^{\circ} - \theta) = \cos (55^{\circ} - \theta)$ and $\theta \leq 115^{\circ}$ we see that $\cos (\theta - 5^{\circ} ) = \cos (55^{\circ} - \theta) \Rightarrow \theta = 30^{\circ}$. Claim 2: $DB = CE$ Proof: Finally, consider $\triangle ACE$. Using sine rule we see that, $CE = AC\frac{\sin 50^{\circ}}{\sin 55^{\circ}}$ (Substituting the value of $\theta = 30^{\circ}$). And, $CE = 2a \sin 55^{\circ}\frac{\sin 50^{\circ}}{\sin 55^{\circ}} = 2a \sin 50^{\circ} $ comparing with (1) we have the desired claim.$\blacksquare$

Easy Let $S$ be on $BA$ such that $\angle ACS = \angle 35$ and $\angle BCS = \angle 40$. Now we have $BSC$ and $CAS$ are isosceles so $DC = BC = SC = SA$ so $C$ is center of circle $BSD$ so $\angle SDB = \angle SCB/2 = \angle 20$ so $\angle SDC = \angle 60 = \angle SCD$ so $DCS$ is equilateral. $SA = SD = SC$ so $S$ is center of $ADC$. $SE = SC = CB = CD$ and $\angle ESC = \angle 100 = \angle BCD$ so triangles $ESC$ and $BCD$ are congruent so $BD = EC$. we're Done.

Welcome to the dark-side of geometry. Here is a trigonometric solution which is possible to find in exam condition? Solution: Let $X$ be circumcenter of $\odot(DAC)$. We will show that $X$ lies on the line $BA$ which is the main part of the problem. [asy][asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(12); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 1.5, xmax = 8.7, ymin = -1.3531765627068162, ymax = 5.9; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); draw(arc((5.54,4.36),0.3963908402195337,-103.16635469383932,-73.16635469383934)--(5.54,4.36)--cycle, linewidth(1) + qqwuqq); draw((4.376061095940733,-0.6156194477431225)--(5.54,4.36)--(8.08,2.7)--(6.886285501273738,-0.08966766757811273)--cycle, linewidth(1) + ffxfqq); draw((5.067219682398144,2.338953246973691)--(6.886285501273738,-0.08966766757811273)--(8.08,2.7)--cycle, linewidth(1) + red); draw(arc((4.376061095940733,-0.6156194477431225),0.3963908402195337,11.833645306160655,41.83364530616067)--(4.376061095940733,-0.6156194477431225)--cycle, linewidth(1) + qqwuqq); draw(arc((5.067219682398144,2.338953246973691),0.3963908402195337,-53.16635469383933,6.833645306160734)--(5.067219682398144,2.338953246973691)--cycle, linewidth(1) + qqwuqq); draw(arc((8.08,2.7),0.3963908402195337,146.83364530616066,186.83364530616072)--(8.08,2.7)--cycle, linewidth(1) + qqwuqq); draw(arc((4.188494228955592,5.2432675511550055),0.3963908402195337,-73.16635469383928,-33.166354693839274)--(4.188494228955592,5.2432675511550055)--cycle, linewidth(1) + qqwuqq); draw(arc((5.54,4.36),0.3963908402195337,-73.16635469383932,-33.16635469383933)--(5.54,4.36)--cycle, linewidth(1) + qqwuqq); draw(arc((6.886285501273738,-0.08966766757811273),0.3963908402195337,66.83364530616068,106.83364530616069)--(6.886285501273738,-0.08966766757811273)--cycle, linewidth(1) + qqwuqq); /* draw figures */ draw((5.54,4.36)--(8.08,2.7), linewidth(1) + red); draw((8.08,2.7)--(6.886285501273738,-0.08966766757811273), linewidth(1) + ffxfqq); draw((6.886285501273738,-0.08966766757811273)--(4.376061095940733,-0.6156194477431225), linewidth(1) + ffxfqq); draw((8.08,2.7)--(4.376061095940733,-0.6156194477431225), linewidth(1) + green); draw((5.54,4.36)--(6.886285501273738,-0.08966766757811273), linewidth(1) + blue); draw(circle((5.067219682398144,2.338953246973693), 3.0343368303469527), linewidth(1) + yqqqyq); draw((5.54,4.36)--(4.188494228955592,5.2432675511550055), linewidth(1) + blue); draw((4.188494228955592,5.2432675511550055)--(4.376061095940733,-0.6156194477431225), linewidth(1) + green); draw((5.067219682398144,2.338953246973691)--(6.886285501273738,-0.08966766757811273), linewidth(1) + red); draw((6.886285501273738,-0.08966766757811273)--(8.08,2.7), linewidth(1) + red); draw((8.08,2.7)--(5.067219682398144,2.338953246973691), linewidth(1) + red); draw((4.188494228955592,5.2432675511550055)--(5.067219682398144,2.338953246973691), linewidth(1) + red); draw((4.376061095940733,-0.6156194477431225)--(5.067219682398144,2.338953246973691), linewidth(1) + red); draw((5.067219682398144,2.338953246973691)--(5.54,4.36), linewidth(1) + ffxfqq); /* dots and labels */ dot((5.54,4.36),dotstyle); label("$B$", (5.591511672807832,4.486981816527643), NE * labelscalefactor); dot((8.08,2.7),dotstyle); label("$C$", (8.326608470322615,2.6635839515177895), NE * labelscalefactor); dot((6.886285501273738,-0.08966766757811273),dotstyle); label("$D$", (6.992092641583517,-0.3754124901653004), NE * labelscalefactor); label("$30^\circ$", (5.366890196683429,3.482791687971492), NE * labelscalefactor,qqwuqq); dot((4.376061095940733,-0.6156194477431225),linewidth(4.pt) + dotstyle); label("$A$", (4.190930704032146,-0.9435726944799651), NE * labelscalefactor); dot((4.188494228955592,5.2432675511550055),linewidth(4.pt) + dotstyle); label("$E$", (4.058800423958968,5.451532861061842), NE * labelscalefactor); dot((5.067219682398144,2.338953246973691),linewidth(4.pt) + dotstyle); label("$X$", (4.59,2.214340999268985), NE * labelscalefactor); label("$30^\circ$", (4.851582104398036,-0.44147763020188935), NE * labelscalefactor,qqwuqq); label("$60^\circ$", (5.472594420741972,2.016145579159218), NE * labelscalefactor,qqwuqq); label("$40^\circ$", (6.793897221473751,2.756075147569014), NE * labelscalefactor,qqwuqq); label("$40^\circ$", (4.415552180156548,4.592686040586186), NE * labelscalefactor,qqwuqq); label("$40^\circ$", (5.776494064910281,3.7206261921032118), NE * labelscalefactor,qqwuqq); label("$40^\circ$", (6.793897221473751,0.5627124983542621), NE * labelscalefactor,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Observe that since $\triangle BCD$ is isosceles, we have $\angle DBC = 40 ^\circ$. Finally apply angle sum property in $\triangle ABC$ to get $\angle CAB = 35 ^\circ$. Call $\angle DAC = \theta$. We will use trigonometry now to prove $\theta = 30^\circ$. Apply Law of Sines in $\triangle ACD$ to get \begin{align} \frac{AD}{\sin(25^\circ)} & = \frac{CD}{\sin(\theta)} \tag{1} \end{align}Apply Law of Sines in $\triangle BAD$ using the fact that $BD = 2\cdot CD\cdot \cos(40^\circ)$ to get \begin{align} \frac{AD}{\sin(30^\circ)} & = \frac{2\cdot CD\cdot\cos(40^\circ)}{\sin(35^\circ + \theta)} \tag{2} \end{align}Now eliminate $AD$ and $CD$ from $(1)$ and $(2)$ by dividing them. This would give \[\frac{\sin(25^\circ)}{\sin(\theta)} = \frac{\cos(40^\circ)}{\sin(35^\circ + \theta)} = \frac{\sin(50^\circ)}{\sin(35^\circ + \theta)}\]From here on, it is just a brute force calculation. Write $\sin(50^\circ) = 2\sin(25^\circ)\cos(25^\circ)$ and expand $\sin(35^\circ + \theta)$ to get \[2\sin(\theta)\cos(25^\circ) = \sin(35^\circ)\cos(\theta) + \sin(\theta)\cos(35^\circ)\]Divide the equation by $\sin(\theta)$ and solve for $\cot(\theta)$ to finally get \begin{align} \cot(\theta) = \frac{2\cos(25^\circ)-\cos(35^\circ)}{\sin(35^\circ)} \tag{3} \end{align}Now we prove the following lemma. Lemma: $\displaystyle\frac{2\cos(25^\circ)-\cos(35^\circ)}{\sin(35^\circ)} = \sqrt{3}$. Proof: The idea is to apply sum to product formula once and then write everything in terms of $30^\circ$ and $5^\circ$ angles and force some cancellations. \begin{align*} \frac{2\cos(25^\circ)-\cos(35^\circ)}{\sin(35^\circ)} & = \frac{\cos(25^\circ)+(\cos(25^\circ)-\cos(35^\circ))}{\sin(35^\circ)} \\ & = \frac{\cos(25^\circ)+\sin(5^\circ)}{\sin(35^\circ)} \\ & = \frac{\cos(30^\circ - 5^\circ)+\sin(5^\circ)}{\sin(30^\circ-5^\circ)} \\ & = \frac{\sqrt{3}\cos(5^\circ)+3\sin(5^\circ)}{\cos(5^\circ)+\sqrt{3}\sin(5^\circ)} \\ & = \sqrt{3} \end{align*}and the lemma is proven.$\square$ Apply the proven Lemma in $(3)$ to get $\cot(\theta) = \sqrt{3} \implies \theta = 30^\circ$ since $0^\circ < \theta < 180^\circ$. Its about time we get back to geometry. Re-define $X$ to the intersection of perpendicular bisector of $\overline{AD}$ and $AB$. Since $\angle DAX = 55^\circ$ we get $\angle AFD = 50^\circ$. By inscribed angle theorem, $X$ must be the circumcenter of $\odot(EDCA)$ restoring the original definition. Applying inscribed angle theorem, we can get that $\triangle XCD$ is equilateral. Due to the equilateral triangle, we would have \[\overline{XE} = \overline{XC} = \overline{XD} = \overline{DC} = \overline{BC}\]This would even give $\angle XEC = \angle XCE = 40^\circ$. Due to all this, one can conclude $\triangle CDB \cong \triangle XEC$ with AAS congruency criteria. It will give $\overline{BD} = \overline{CE}$ and we are done. $\blacksquare$

Let $\angle DAC = \theta$ and therefore $\angle BDA = 115 - \theta$. By quadrilateral Ceva, we have that: $$ \frac{\sin \theta}{\sin(115 - \theta)} = \frac{\sin35}{\sin30} \frac{\sin40}{\sin75} \frac{\sin25}{\sin40}=\frac{\sin25 \sin35 \sin85}{\sin75 \sin30 \sin85} = \frac{\left(\frac{\sin75}{4}\right)}{\sin75 \sin30 \sin85} = \frac{\sin30}{\sin85}$$ where the last step is due to the identity: $\sin(60 - x)\sin(x)\sin(60 + x) = \frac{\sin(3x)}{4}$. Thus we cleanly get $\theta = 30^\circ$ and now by sine rule, we have that: $$BD = CD \cdot \frac{\sin100}{\sin40} = CD \cdot \frac{2\sin50\cos50}{\sin40} = CD \cdot \frac{\sin50}{\sin30} = CE $$ And done! P.S. refer to diagram above