Hint : reflect points B , D wrt midpoint of CE

Let $D' \neq D$ be a point on $DM$ such that $DM=D'M.$ Then $OD=OD'$ and hence $B,C,D,D'$ are cyclic. Let $B'$ be a point on $[BC$ such that $B'C=AE.$ Since $AE$ and $BC$ are parallel to each other, we conclude that $ACB'E$ is a parallelogram. Then $A,M,B'$ are collinear. Since $BB'=AB$ and $AM=MB'$ we conclude that $BM$ and $AB'$ are perpendicular. Let $\angle B'BM = \angle MBA= \alpha.$ $AD'B'D$ is also a parallelogram and therefore $AD=B'D'.$ Let $\angle D'B'B = q$ and $\angle DBM=p.$ Then since $\angle AED'=\angle B'CD=\angle ED'C-\angle D'CB$ we get $\angle DCD'=\angle DBD'=180- 2\alpha.$ Hence $\angle B'BD'=180-\alpha-p$ and $\angle DBA=\alpha +p.$ $\angle DAB=180-2\alpha-q$ and therefore $\angle ADB=\alpha + q-p.$ We shall show that $p=q.$ By sine theorem in triangles $ADB$ and $B'D'B$ we get $\frac{\sin (\alpha+q-p)}{\sin (\alpha +p)} = \frac{AB}{AD}=\frac{BB'}{B'D'} = \frac{\sin (\alpha+p-q)}{\sin (\alpha+p)}$ So, we get $\sin (\alpha+q-p) = \sin (\alpha+p-q)$ and since $\alpha <90$ we get $p=q.$ So, we are done.

As skytin suggested, let $B',D'$ be the reflections of $B,D$ with respect to $M$. Then $OM$ becomes the perpendicular bisector of $DD'$ meaning $OD=OD'$, so that $D'$ lies on $(BCD)$. Note that $M$ is the midpoint of $BB'$ and $CE$ so $B'EBC$ is a parallelogram, implying $B'E||BC$. Then $B'E||AE$ so that $B',E,A$ are collinear. Let $T\in (AB)$ be such that $AT=AE$. Then given the condition $AB=AE+AC$ this implies $BT=BC=EB'$ and clearly $\triangle ABB'$ is isosceles. Now by a simple angle chase, $\angle D'ED=\angle D'CD=\angle D'BD=\angle D'B'D$, i.e. $D'EB'D$ is cyclic. Then $\angle ED'D=180^{\circ}-\angle EB'D$. But $\angle ABC=\angle CDE\implies\angle ABD+\angle CBD=\angle CDD'+\angle EDD'$. But $\angle EDD'=\angle CD'D=\angle CBD$ hence $\angle ABD=\angle CDD'=\angle ED'D=180^{\circ}-\angle EB'D$. Thus, $ABDB'$ is cyclic. Then $\angle BDA=\angle BB'A=\angle ABB'$. Now $\angle ABC=\angle ABB'+\angle CBB'=\angle ABB'+\angle AB'B=2\angle ABB'=2\angle BDA$. So $2\angle BDA=\angle ABC=\angle CDE$.

Actually, it was really fun to design the construction on geometry software. Here is how to do it (I use CarMeTal, for those interested):

I think the property of a diagram being constructible with solely a compass and a straight edge is a tacit rule among ISL geometry problems, and even though this is constructed with software, each step is entirely possible to do on paper with the compass and straight edge (along with a sharp pencil!). For example, when I said to reflect $B$ in the line $AC$, just draw the perpendicular from $B$ to the line $AC$ and then draw in the circle with the foot of this perpendicular as the centre, with $B$ on the circumference. Then $B$'s antipode is $B_1$.

Throughout my solution I will use directed angles modulo $180^\circ$. As in the above solutions, let $B'$ and $D'$ be the reflections of $B$ and $D$ with respect to $M$, respectively. We find that $AB'=AE+EB'=AE+BC=AB$, hence $\triangle ABB'$ is isosceles. Hence $\angle ABB'=\angle BB'A=\angle B'BC$, so $BB'$ is the angular bisector of $\angle ABC$ and $\angle BB'A=\angle ABB'=\tfrac12 \angle ABC=\tfrac12 \angle CDE$. Now we are going to prove that the quadrilateral $ABDB'$ is cyclic, because that would yield $2\angle BDA=2\angle BB'A=\angle CDE$, as we have to prove. Because $\triangle OMD \simeq \triangle OMD'$, we get $|OD|=|OD'|$ and hence $BCDD'$ is cyclic. Let $F$ be the intersection of the lines $AB$ en $ED'$. Then $\angle FBC=\angle ABC=\angle CDE=\angle ED'C=\angle FD'C$, so $FBCD'$ is also a cyclic quadrilateral. And so $B,C,D,D'$ and $F$ all lie on one circle. Furthermore, since $B'ED'D$ is the reflection of $BCDD'$ with respect to $M$, that quadrilateral is also cyclic. Now we get: $\angle AB'D=\angle EB'D=\angle ED'D=\angle FD'D=\angle FBD=\angle ABD$ and so $ABDB'$ is cyclic and we are done. By introducing the point $F$, the condition $\angle ABC=\angle CDE$ becomes equivalent with $F$ lying on a certain circle. So in this way you've got rid of all the ugly conditions, and therefore you get a much easier angle chase

Here is my solution for this nice G5 : Let a point on such that . Otherwise, let us consider the homotethy and a point such that . Also, i give below a litte Lemma wich is trivial but i will use it a lot of times. Lemma: Let a quadrilatere, this one is a parallelogram if and only if his diagonals meet each other in her midpoints. Since and , and is the midpoint of , we have is a parallelogram and is the midpoint of . Otherwise, since we must prove that : because the triangle is iscoscele. And so we are left to prove that points are concyclic. First, it's not hard to see that since and is the midpoint of and then the triangle is iscoscele, hence points are concyclic. Even since is the common midpoint of the we get that are parallelograms. Hence : . Moreover, since and and hence which give us : . But we have which means, i.e : . Hence, , which is obviously true, and hence the points are concyclic and hence we get the desired result which is :. Q.E.D

Suppose $B=0,C=1,A=a,E=e,D=d,M=m$ with $im(a)=im(e)$.Suppose $\odot{BCD}=X$ and so $X.\overline{X}=(X-d)(\overline{X-d})$.That implies $\overline{d}X+d(1-x)=d\overline{d}$. Also from $X\overline{X}=(X-1)(\overline{X-1})$ we've $X+\overline{X}=1$. So solving we get $X=\frac{d\overline{d}-d}{\overline{d}-d}$. Now from the condition $\angle{ABC}=\angle{CDE}$ we get $\frac{a(e-d)}{d-1}\in\mathbb R$ and basically so, $\frac{a(e-1)}{d-1}\in\mathbb R\implies a(e-1)(\overline{d}-1)\in\mathbb R$. Now $m=\frac{e+1}{2}$.Also as $\angle{DMO}=\frac{\pi}{2}$ so, $re(\frac{X-m}{d-m})=0$. Now take $a=x+iy,e=m+iy,d=p+iq$. From the condition $ a(e-1)(\overline{d}-1)\in\mathbb R$ directly we get $q(x(m-1)-y^2)=y(p-1)(x+m-1)$. Now from the condition $re(\frac{X-m}{d-m})=0$ directly we've $qm(2p-m-1)=(2q-y)(p^2+q^2-p-qy)$. Now just eliminating $m$ from above two expressions we get $2y(2py-px-pq)(p^2-px-y^2+qy)=x((p^2-px-y^2+qy)^2+(2py-px-pq)^2)$. Here we're asked to show $2\angle{BDA}=\angle{CDE}$. So we need to show $2arg(\frac{d}{d-a})=arg(a)$.Suppose $arg(a)=\theta \implies tan(\theta)=\frac{y}{x}$. If $arg(\frac{d}{d-a})=\alpha$ then $tan(2\alpha)=\frac{2(2py-px-pq)(p^2-px-y^2+qy)}{(p^2-px-y^2+qy)^2+(2py-px-pq)^2}$ , which is indeed true, so $\theta=2\alpha$ , that's what all we needed to show,so we're done.

Let $X$ be the midpoint of $CD$.Then $OX \perp CD$ and $OM \perp MD$,so $OMDX$ is cyclic.Thus $\angle{CBD}=\angle{DOX}=\angle{DMX}=\angle{MDE}$ since $MX \parallel DE$.Let $Y$ be the midpoint of $AB$.Then it is well known that $MY=\frac{AB+AE}{2}=MA=MB$ so $Y$ is the circumcenter of $AMB$.Consequently from straightforward angle chasing we get that $MA,MB$ are the internal angle bisectors of $\angle{EAB}$ and $\angle{ABC}$ respectively.Now applying sine rule in $\triangle{AMD}$ and $\triangle{BMD}$ we get $\frac{AM}{DM}=\frac{sin(D+E-x-y)}{-sin(A/2+E+y)}$ $\frac{BM}{DM}=\frac{sinE}{cos(A/2+D+E-x)}$ where $\angle{BDC}=x,\angle{ADE}=y$.Dividing the expressions and using the fact that $\frac{BM}{AM}=tan\frac{A}{2}$ we get $cos(x+y)=cos\frac{D}{2}$.Thus $2\angle{ADB}=\angle{EDC}$,as desired.

A very nice problem indeed, with some beautiful constructions... Here, Reflect $B,D$ about $M$ to get $L,K$ respectively. Now observe that $OK=OD=OB=OC$ implies that $B,K,D,C$ are concyclic. Now we have by the length conditions, $A,E,L$ are collinear and $AB=AL$. Now notice that $\angle BDL= \angle BKL =\angle BKC+\angle CKL=\angle BDC+\angle BDE=\angle CDE=\angle ABC=180-\angle BAL $ This implies that points $B,A,L,D$ are concyclic and since $BA=AL$ by Fact 5 we have $\angle BDA=1/2\angle CDE$. This completes the proof.

anantmudgal09 wrote: A very nice problem indeed, with some beautiful constructions... Here, Reflect $B,D$ about $M$ to get $L,K$ respectively. Now observe that $OK=OD=OB=OC$ implies that $B,K,D,C$ are concyclic. Now we have by the length conditions, $A,E,L$ are collinear and $AB=AL$. Now notice that $\angle BDL= \angle BKL =\angle BKC+\angle CKL=\angle BDC+\angle BDE=\angle CDE=\angle ABC=180-\angle BAL $ This implies that points $B,A,L,D$ are concyclic and since $BA=AL$ by Fact 5 we have $\angle BDA=1/2\angle CDE$. This completes the proof. Very good!

skytin wrote: Hint :reflect points B , D wrt midpoint of CE anantmudgal09 wrote: Reflect $B,D$ about $M$ to get $L,K$ respectively. Can anybody explain how you're supposed to come up with the idea of reflecting ? How you even draw one example of a pentagon satisfying all these contrived conditions with ruler and compass (before having the idea to reflect)? (I don't understand post #4 properly, i.e why should $D \in \omega_{B_2}(A,B)$ ?)

Nice construction problem! First, construct point $X$ on $\overline{AB}$ such that $AE=AX$ and $BX=BC$. Then observe that $\angle CXE=90^{\circ}$ so $BM,CM$ are perpendicular bisectors of $CX,AX$ respectively. Hence $BM,CM$ bisects $\angle ABC$ and $\angle BAE$ respectively. Draw parallelogram $CDEP$. From $\angle DMO=90^{\circ}$, we get $P\in\odot(BCD)=\omega$. Moreover, if $Q=EP\cap AB$, then notice that \begin{align*} \angle EAQ &= 180^{\circ}-\angle ABC \\ &= 180^{\circ}-\angle CDE \\ &= \angle DCP \\ &= \angle DBP \\[4pt] \angle AEP &= 180^{\circ}-\angle BCD = \angle BPD \end{align*}So $\triangle AEQ\sim\triangle BPD$ which means $\angle AQE=\angle BDP$ or $Q\in\omega$. Now $\angle DQE = \angle DBP = \angle QAE$ and $\angle DEQ = \angle DCP = \angle DBP$ or $DQ,DE$ are tangents to $\odot(QAE)$. Let $M$ be the midpoint of $EQ$. Then by symmedian lemma, $\angle DAQ = \angle MAE = \angle MBP$. Finally, angle chase \begin{align*} \angle BDA &= 180^{\circ} - \angle DBQ - \angle DAQ \\ &= \angle CBP - \angle MBP \qquad\qquad (\because CP=DQ)\\ &= \angle CBM \\ &= 0.5\angle CDE \end{align*}so we are done.

Let $N, K, L$ be the midpoint of $BD$, $AB$ and $CD$, respectively. $\angle{OMD}=\angle{OND}=\angle{OLD}=90^\circ$, So $ONLDM$ is cyclic. Since $AE//MK//BC$, $MK=\displaystyle\frac{AE+BC}{2}=\displaystyle\frac{AB}{2}=KA=KB$, $\angle{AMB}=90^\circ$. Then $\angle{MKB}=\angle{EAB}=180^\circ-\angle{EDC}=\angle{MLD}=\angle{MND}$ So $KBNM$ is cyclic. $\therefore 2\angle{ADB}=2\angle{KNB}=\angle{MNB}=180^\circ-\angle{MND}=\angle{EDC}$

Wait is this a G5???This is pure angel chase [asy][asy] size(8cm); pair A,B,D,P,E,M,C,O,X,T; A=dir(215); B=dir(305); P=dir(120); E=0.2*P+0.8*A; M=midpoint(P--B); C=2*M-E; D=dir(100); T=2*M-D; O=circumcenter(B,C,D); X=foot(O,D,M); T=2*X-D; filldraw(E--P--D--cycle,gray); filldraw(C--B--T--cycle,gray); draw(A--B--C--D--P--cycle,linewidth(1.2)); draw(unitcircle); draw(B--P); draw(E--C); draw(B--D); draw(D--T); draw(E--D); draw(T--C); draw(T--B); draw(circumcircle(B,T,C),dashed); dot("$A$",A,dir(200)); dot("$B$",B,dir(-30)); dot("$C$",C,dir(0)); dot("$D$",D,dir(80)); dot("$E$",E,dir(180)); dot("$P$",P,dir(120)); dot("$M$",M,2*dir(240)); dot("$T$",T,dir(270)); [/asy][/asy] Select $P$ on $\overrightarrow{AE}$ such that $AP=AB$.Now by the problem stipulation we have that $PEBC$ is parallelogram.So $P$ lies on $\overline{MB}$.Because $\overline{BC} \parallel \overline{AP}$ we have $\angle PAB = 180^{\circ} - \angle ABC = 180^{\circ} - \angle CDE$ and so $2\angle APB = \angle CDE$.Whence it suffices to show $\angle APB = \angle ADB$ or $PABD$ is cyclic. Now let $T$ be the reflection of $D$ about $M$.By the problem condition we know $(DCBT)$ is cyclic.Whence $\angle BDC = \angle BTC$.Now a simple claim: Claim: We have $\triangle PDE \cong \triangle BTC$ Proof: Notice that $\triangle PDE$ is the reflection of $\triangle BTC$ about $M$ Now by the claim $\angle BDC = \angle BTC = \angle PDE$ and whence $\angle EDC = \angle PDB$.Now compute: $$\angle PAB + \angle PDB = \angle PAB + \angle EDC = \angle PAB + \angle ABD = 180^{\circ}$$because $\overline{BC} \parallel \overline{AP}$.Whence $(PABD)$ is cyclic and we are done $\blacksquare$

char2539 wrote: Wait is this a G5???This is pure angel chase Nice solution! Angel chasing is my favourite hobby, too!

Let $F$ be the reflection of $B$ wrt $M$. As $AE\parallel BC\parallel EF$, we get that $A,E,F$ are collinear and thus, $AF=AE+EF=AE+BC=AB$. Let $X$ be the midpoint of $BD$, $Y$ be the midpoint of $CD$. Thus, as $\angle DMO=90^{\circ}=\angle DXO=\angle DYO$, we have $DXYMO$ cyclic quadrilateral. Hence, $$\measuredangle EDC=\measuredangle MYC=\measuredangle MYD=\measuredangle MXD=\measuredangle MXB=\measuredangle FDB$$and as $$\measuredangle EDC=\measuredangle CBA=\measuredangle FBA+\measuredangle AFB=\measuredangle FAB$$and hence we conclude that $AFDB$ is a cyclic quadrilateral. Thus, $$2\measuredangle ADB=2\measuredangle AFB=\measuredangle CBA=\measuredangle EDC,$$we are done.

Reflect $B$ across $M$ to $B'$. Let $X$ be a point on $AB$ such that $AX = AE$. By the given length condition, $BX = BC$ as well. Since $\angle B'EC = \angle ECB = 180 - \angle AEC$, $A,E,B'$ are collinear. Also, $\angle EXC = 180 - \angle EXA - \angle CXB = 180 - (90 - \frac{\angle EAX}{2}) - (90 - \frac{\angle XBC}{2}) = 90^\circ$. So, since $M$ is midpoint of hypotenuse in $CEX$, $MX = MC$. Since $BX = BC$, this means $BM \perp CX$ Let $Y,Z$ be midpoints of $BD$ and $CD$. Since we're given $\angle DMO = 90^\circ$, this means $DMOYZ$ is cyclic. Since $Y,Z$ are midpoints, there is a homothety centered at $D$ with ratio $2$ that takes $(DYZ)$ to $(DBC)$ and so the reflection of $D$ across $M$, call it $D'$, lies on $(BCD)$ Since this means $\angle BD'C = \angle BDC$, reflecting stuff across $M$, we have that $\angle B'DE = \angle BDC \implies \angle BDB' = \angle EDC = \angle ABC = 180 - \angle B'AB$ and so $ABDB'$ is cyclic. So, $2 \angle BDA = 2 \angle AB'B = 2 \angle ABB' = \angle ABC = \angle CDE$ and so we are done. $\blacksquare$

Let $B'$ and $D'$ be the reflections of $B$ and $D$ over $M.$ Notice $B$ lies on $\overline{AE}$ as $BCB'E$ is a parallelogram and $D'$ lies on $(BCD)$ as $\triangle OMD\cong\triangle OMD'.$ Also, the length condition implies $AB=AB'.$ Claim: $ABDB'$ is cyclic. Proof. Notice $B,'E,D,'$ and $D,$ are the reflections of $B,C,D,$ and $C'$ over $M,$ respectively; hence, $B'ED'D$ is cyclic. Since $CDED'$ is a parallelogram, \begin{align*}\measuredangle ABD&=\measuredangle ABC-\measuredangle DBC=\measuredangle ABC-DD'C\\&=\measuredangle CDE-D'DE=\measuredangle CDD'=\measuredangle ED'D=\measuredangle EBD.\end{align*}$\blacksquare$ We see $$\measuredangle CDE=\measuredangle ABC=\measuredangle ABB'+\measuredangle B'BC=\measuredangle BDA+\measuredangle BB'A=2\measuredangle BDA.$$$\square$

Let $B'$ and $D'$ be the reflections of $B$ and $D$ across $M$. We have $\angle DMO=\angle D'MO$, so $OD=OD'$. Thus, $BCDD'$ is a cyclic trapezoid, and $EB'DD'$ is also a cyclic trapezoid which is the reflection across $DD'$. $~$ By symmetry, $B'E\parallel BC\parallel AE$ so $A,E,B'$ collinear. \[\angle BDB'=\angle CDE=\angle ABC=180^\circ-\angle BAE=180^\circ-\angle BAB'\]so $ABDB'$ is cyclic. Note that $AB=BC+AE=B'E+AE=AB'$ so \[\angle ADB=\angle AB'B=90^\circ-\frac12 \angle B'AB=\frac12 \angle BDB'=\frac12 \angle CDE\]as desired.

Is there a solution to this problem by constructing $F = \overline{BC} \cap \overline{AM}$ (such that triangle $FBA$ is isosceles) and then proving the two claims $F, N, M, A$ collinear, where $N$ is the midpoint of $\overline{CD}$, and $F, O, D'$ collinear, where $D'$ is the reflection of $D$ over $M$? This was the most intuitive way to construct the diagram for me, and both claims seem to be true. They together also imply the problem after an angle chase (using $ONDM$ cyclic.) Unfortunately, I wasn't able to make much headway proving either claim, but they intuitively do not feel hard to prove. (Thus I suspect that they're actually wrong -- which might make one of the funniest coincidences in geometry.)