tag:blogger.com,1999:blog-6933544261975483399.comments2019-08-11T04:42:47.812-07:00Go GeometryAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5812125tag:blogger.com,1999:blog-6933544261975483399.post-66685459090547782022019-08-10T17:06:48.677-07:002019-08-10T17:06:48.677-07:00Yes ! ThanksYes !<br />ThanksAntonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-38903841894116601852019-08-10T07:22:13.751-07:002019-08-10T07:22:13.751-07:00do you accept a solution applying homothety?do you accept a solution applying homothety?Gewürtztraminer68https://www.blogger.com/profile/00329980114313015297noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-81564703131434756502019-08-07T07:52:57.101-07:002019-08-07T07:52:57.101-07:00This is a link to the drawing ΔABC is isosceles on...<a title="Link to the drawing" href="https://drive.google.com/file/d/16KGzDMJ1l4tyTkGsBEKF-jvHVQkT8v0m/view?usp=sharing" rel="nofollow"> This is a link to the drawing</a><br />ΔABC is isosceles on B =&gt; x=7α<br /><br />Define A’ symmetric A by BD<br />Then DA=DA’, BA=BA’, ang(BDA) = 30° =&gt; ang(BDA′) = 30° =&gt; ang(ADA′) = 60°<br /><br />DA=DA’ and ang(ADA′) = 60° =&gt; ΔADA’ is equilateral<br /><br />ΔABC is congruent to ΔABA’ (SSS) =&gt; ang(BAA′) = ang(BA&#39;A) = x<br />Therefore ang(DAA′) = x +5α =60°<br />But x=7α =&gt; 7α + 5α =60 =&gt; 12α =60 =&gt; α=5° =&gt; x= 35°rv.littlemanhttps://www.blogger.com/profile/15820037721044128612noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-83870675374051938152019-08-07T07:42:56.520-07:002019-08-07T07:42:56.520-07:00Draw the circle C1, with incenter D and radius DA=...Draw the circle C1, with incenter D and radius DA=r1<br />Draw the circle C2, passing through D, B and C, with incenter 0 and radius DO=r2 : OD=OB=OC<br />Let suppose O is not on AC, neither on C1<br />ang(BCD) = 30° intercepts arc DB =&gt; ang(DOB) = 60°<br />ang(DOB) = 60° , OD=OB =&gt; ΔDOB is equilateral<br />Therefore DO=DB=r1 =&gt; O is on C1<br /><br />ΔDOB is equilateral =&gt; DO=DB and DB=DA =&gt; DO=DA<br />O is on C1 =&gt; ang(AOB) = 90°<br />ang(AOB) = 90° , ang(BOD) = 60° =&gt; ang(DOA) = 30°<br /><br />Define E, the intersection of C1 and AC. ang(AEB) = 90<br />DA=DB=DE therefore ΔADE is isosceles in D =&gt; ang(AED) = 2x<br />ang(DAE) = 2x and ang(DCA) = x =&gt; ang(CDB) = 3x <br />ang(EDB) = ang(DAE) + ang(DEA) =4x<br />ang(EDC) = ang(EDB) - ang(CDB) = 4x-3x = x <br />ang(EDC) = x and ang(ECD) = x therefore ΔCED is isosceles in E =&gt; ED=EC<br /><br />O=E?<br />What we know about E : E on C1, on AC, ang(AEB) = 90 and ED=EC<br />What we know about O : O on C1, ang(AOB) = 90 and OD=OC=OB<br /><br />Define circle C3, with incenter C and radius C0=CE=r1<br />It means that E and O are on C1 and C3<br /><br />Let&#39;s examine the various situations with respect to the intersection(s) of C1 and C3:<br />Option 1 : There are only two points, intersection of the two circles C1 and C3 with the same radius r1<br /><br />Option 1a : E≠O : <br />let suppose E and O are on these two different intersections (Cf. graph)<br />OD=OC=ED=EC =&gt; ODEC is a diamond =&gt; DC is the angle bisector of ang(EDO) and ang(ECO)<br />ang(EDC)=x =&gt; ang(CDO)=x<br />ang(CDB)=3x, ang(CDO)=x =&gt; ang(ODB)=2x<br />Since ΔDOB is equilateral =&gt; 2x=60 =&gt; x=30<br /><br />Option 1b : E=O <br />If O=E, then ang(DEA) = ang(DOA), 2x=30 =&gt; x=15°<br /><br />Option 2 : There are only one point, intersection of the two circles C1 and C3<br />Then E=O Cf. Option 1b<br /><br />Option 3 : There is no point, intersection of the two circles C1 and C3<br />Since EC=ED =&gt; E belongs to C1 and C3 : this option is impossible<br /><br />Conclusion : 2 solutions, x=15° and x=30°. See graph<br /><br /><a title="Link to the drawings" href="https://drive.google.com/file/d/1dcEYkeahFpG4C520MyXtwg29XmGyna0P/view?usp=sharing" rel="nofollow"><br /> This is a link to the drawings<br /></a>rv.littlemanhttps://www.blogger.com/profile/15820037721044128612noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-27257384268409593202019-08-07T07:26:49.932-07:002019-08-07T07:26:49.932-07:00If x=20 then it is possible to draw a triangle ABC...If x=20 then it is possible to draw a triangle ABC such as Ang(A)=2x, Ang(ACD)=x, Ang(DCB)=30.<br />But, in that case, DA will not be equal to DB…<br /><a title="Link to the drawing" href="https://drive.google.com/file/d/1gcMOMxncS5RImt8FtdoFVb0PkHt7H0zu/view?usp=sharing" rel="nofollow">See it in this drawing<br /></a>rv.littlemanhttps://www.blogger.com/profile/15820037721044128612noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-17867577211572752152019-08-07T05:43:23.730-07:002019-08-07T05:43:23.730-07:00Hello Sumith, My finding is that B cannot be a rig...Hello Sumith,<br />My finding is that B cannot be a right angle. You can check my post below for details.<br />Euclidean regards.<br />GregGreghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-70942778122742844242019-08-06T11:01:50.987-07:002019-08-06T11:01:50.987-07:00https://photos.app.goo.gl/oBvDsrGZ2wwrWe9V7 let B...https://photos.app.goo.gl/oBvDsrGZ2wwrWe9V7<br /><br />let BI meet AC at N and arc AC at D<br />Let CI meet arc AB at M<br />We have I is the midpoint of BD<br />and Triangle DIC is isoceles<br />Triangle DNC simillar to DCB ( case AA)<br />so DC^2=DN.DB=2DN.DI=2DN.DC<br />so DN/DC=DN/DI=1/2 =&gt; N is the midpoint of ID<br />and NI/NB=1/3<br />Area(AIC)/Area(ABC)= NI/NB= 1/3Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-57809427743297033362019-08-06T05:35:36.524-07:002019-08-06T05:35:36.524-07:00How to find the inradius of tr. ABC? How to find the inradius of tr. ABC?<br /> Unknownhttps://www.blogger.com/profile/15012247681944808200noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-7724858489836313162019-08-05T15:00:41.346-07:002019-08-05T15:00:41.346-07:00Let BI extended meet AC at D and circle O at E &l...Let BI extended meet AC at D and circle O at E<br /><br />&lt; EAI = EIA = (A+B)/2 so AE = EI = IB since OI is perpendicular to BE<br /><br />Further, since &lt; EAD = &lt; ABE = B/2, Tr.s ADE &amp; ABE are similar and so<br />AD/c = AE/BE = 1/2 so that AD = c/2<br /><br />But AD = bc/(a+c), hence c/2 = bc/(a+c) from which, a+c= 2b<br /><br />Now S1 = rb/2 = (S/s)*(b/2) = Sb/(a+b+c) = Sb/(3b) = S/3<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-55429819334921987592019-08-03T06:19:44.413-07:002019-08-03T06:19:44.413-07:00It is interesting to note that if the condition “a...It is interesting to note that if the condition “angle B is an obtuse angle” were to be lifted, which was the case in the original problem statement, the 2 solutions x=15° or x=30°, that Ajit found at the beginning through trigonometry, can be determined in various “Euclidean” ways.<br /><br />1/ In Sumith Peiris first proof : the statement “since a NOT b” is required to get rid of the 0=0 trivial solution to the equation, but one could very well look at what happens when a=b and find that in that case x=30°, which is a valid solution.<br /><br />2/ In Piet van Kampen’s proof : after finding that DFC is equilateral, “ang(FDB)=x; ang(BDC)=3x ang(FDC)=4x” because B is obtuse, but if B were acute, then F would be, on BC, on the other side of DB and the series of equations would be “ang(FDB)=x; ang(BDC)=3x ang(FDC)=2x” which gives x=30°, which is a valid solution.<br />[In the case where B would be allowed to be a right angle, Piet’s proof would not work because D, B and F would be aligned and triangles DFB and DCE would not be equal triangles anymore, so that the above equations would not stand. And in that case, when B is a right angle, one may wonder whether x=20°could be a third valid solution.]<br /><br />3/ Lastly, a different way of looking at the problem could be the following (see animated graph <a title="Animation Solution to Problem 046 by Greg" href="https://drive.google.com/file/d/1vewoBHFaZR3tb-Z-ydWwdq-5DvtCWT7V/view?usp=sharing" rel="nofollow">here</a>).<br /><br />Allow ang(BCD) to take any value, the remaining of the problem’s construction being unchanged, and define E as the foot of the altitude from B : then AD=BD=DE (ABE right triangle in E) and DE=EC because ang(CDE)=pi-x-(pi-2x)=x.<br /><br />Draw circle C1 of diameter AB, and circle C2 of center E and radius EC=ED, and allow E to “move” from A to B on arc(AB), keeping C on AE such that EC=ED=AD : the problem is equivalent to finding the positions of E that make ang(BCD) to be equal to 30°.<br /><br />Let G be the intersection of C1 and C2 on arc(AB) : DEG is equilateral (because DE=DG in C1 and ED=EG in C2) therefore ang(DCG) = ang(DEG)/2 = 30°. So as E “moves” on arc(AB) from A to B, the equilateral triangle DEG rotates around D within C1 and G “moves” on arc(AB) ahead of E towards B.<br />For ang(DCB) to be equal to 30°, as ang(DCG)=30°, B, C and G must be aligned (even allowing G and B to coincide, as the case may be).<br /><br />If G and B do not coincide, since BDG is isosceles in D and CEG is isosceles in E, Ang(DBC)=ang(DBG)=ang(BGD) and ang(ECB)=ang(ECG)=ang(CGE) therefore ang(DBC)+ang(ECB) = ang(BGD)+ang(CGE) = pi -ang(DGE)=2*pi/3 therefore 2x=pi/3 and x=30° (first solution – first stop of the animation).<br /><br />If E is allowed to move further on arc(AB), at some point G becomes equal to B and ang(DCB)=ang(DCG) = 30° which is also a valid configuration. In that case BDE is equilateral and ang(BDE)=4x=60° therefore x=15° (second solution – second stop of the animation).<br />So there are 2 valid solutions to the original problem, x=30° or x=15°.<br /><br />And the answer to the question about x=20° being a valid solution or not (see 2/ above) is : it cannot be because when B, G and C are aligned : ang(ABC)=ang(ABG)=ang(ADG)/2. By our construction, ang(ADG) can only be equal to pi when G=B which is when x=15° and ang(ABC)=105°. And this is confirmed by the attached <a title="B cannot be a right angle" href="https://drive.google.com/file/d/1fMLHKhMmCYNIhZIob-FGI4xucO2jZ7gd/view?usp=sharing" rel="nofollow">graph</a> (simple trigonometry would also confirm it easily).Greghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-44303476261653435772019-07-31T12:47:38.099-07:002019-07-31T12:47:38.099-07:00AD meet BC in E. Let EF parallel with DC, F holds ...AD meet BC in E. Let EF parallel with DC, F holds on BD segment. &lt;BDE=alpha+beta and &lt;DEF=90-alpha-beta. Then FE perpendicular to BD, so DC perpendicular to BD. Then DE=BE=EC. Result AE is bisector and median at the same time, so AB=AC. From symmetry result alpha=x.Palhegyi Laszlohttps://www.blogger.com/profile/18374894376925620852noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-91298649106174044282019-07-31T11:02:49.028-07:002019-07-31T11:02:49.028-07:00Since &lt; BDE = A, &lt; MBG = (90-A) - (90-C) = C...Since &lt; BDE = A, &lt; MBG = (90-A) - (90-C) = C-A = &lt; MCB<br /><br />It follows that BM^2 = MG.MC = 10<br />So BM = sqrt10<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-6674311140710757702019-07-30T09:57:10.913-07:002019-07-30T09:57:10.913-07:00If B is obtuse x = 15 If B is acute x = 30 If B is...If B is obtuse x = 15<br />If B is acute x = 30<br />If B is right x = 20Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-17310059590440932652019-07-28T12:15:48.382-07:002019-07-28T12:15:48.382-07:00Solution 3 Drop a perpendicular BP from B to AC, ...Solution 3<br /><br />Drop a perpendicular BP from B to AC, P on AC. Draw equilateral CDQ on DC.<br /><br />Now AD = DP = PC = DB = BQ and so Tr.s CDP &amp; BDQ are congruent SSS<br /><br />So &lt; BDQ = 60-3x = &lt; PCD = x from which,<br /><br />x = 15<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-12668949378325870802019-07-27T02:26:25.111-07:002019-07-27T02:26:25.111-07:00Sorry Piet, you are right, DFC IS equilateral. The...Sorry Piet, you are right, DFC IS equilateral. The typo &quot;DE=EB&quot; instead of &quot;DE=EC&quot; in the previous sentence of your proof had mislead me in my earlier comment.Greghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-41733776691488488142019-07-21T04:26:39.819-07:002019-07-21T04:26:39.819-07:00why AE//DB？why AE//DB？Unknownhttps://www.blogger.com/profile/10583896946492551271noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-73452281881333676672019-07-20T05:35:31.195-07:002019-07-20T05:35:31.195-07:00If Point&#39;s Q(R,0) and O(0,0) Define: (x-r)²+y...If Point&#39;s<br />Q(R,0) and O(0,0)<br /><br />Define:<br />(x-r)²+y²=R²<br />x²+y²=r²<br /><br />On Point O, can define two lines<br />y=a*x (Line on Point&#39;s OCD) and y=b*x (Line on Point&#39;s OEF)<br /><br />So:<br />(x-R)²+y²=R²<br />y=a*x<br />=&gt; Point D [R/(1+a²),(a*R/(1+a²))]<br /><br />(x-R)²+y²=R²<br />y=b*x<br />=&gt; Point F [R/(1+b²),(b*R/(1+b²))]<br /><br />x²+y²=R²<br />y=a*x<br />=&gt; Point C [r²/(1+a²),(a*r²/(1+a²))]<br /><br />x²+y²=R²<br />y=b*x<br />=&gt; Point B [r²/(1+b²),(b*r²/(1+b²))]<br /><br /><br />Solve for the Slope k (germ. Steigung) at the Line on the Point&#39;s CE<br />k(1)=a-b (i)<br /><br /><br />On Point&#39;s CD can construction a Normal line, who will Intersect the (Big) Circle (-R,0)<br /><br />n: y=-(1/a)*(x-(R+r²)/(2*(1+a²)))+a*(R+r²)/(2(1+a²))<br /><br />=&gt; The Intersect Point&#39;s:<br />G [(p+q)/(2(1+a²)),(a*(r²-R)-p/a)/(2(1+a²))] and H [(q-p)/(2(1+a²)),(a*(r²-R)+p/a)/(2(1+a²))]<br /><br /><br />The same procedure can be define the Point&#39;s M and N.<br />...<br />=&gt; The Intersect Point&#39;s:<br />M [(u+v)/(2(1+b²)),(b*(r²-R)-u/b)/(2(1+b²))] and N [(v-u)/(2(1+b²)),(b*(r²-R)+u/b)/(2(1+b²))]<br /><br />In wich:<br />p=a*√(R²(4a²+3)+r²(2R-r²)) , q=2a²R+R+r²<br /><br />u=b*√(R²(4b²+3)+r²(2R-r²)) , v=2b²R+R+r²<br /><br />__________<br /><br />Now it can define a new Lines with Point&#39;s G and N, M and H.<br />The Slope from the new Lines, a solve will be show:<br /><br />k(2)=a-b, line with Point&#39;s G and N (ii)<br />k(3)=a-b, line with Point&#39;s M and H (iii)<br />__________________________<br /><br />So:<br />For the three Lines, the Slope is a same. <br />A prove for the Parallelity lines.<br /><br /><br /><br /><br /><br /><br /><br />Ludwig Mercknoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-71961959617219279112019-07-14T18:35:42.089-07:002019-07-14T18:35:42.089-07:00From previous problem, A,C,H co line, M,E,B co lin...From previous problem, A,C,H co line, M,E,B co line.<br />Connect ACH, connect BEM. <br />Connect AB; <br />so angle AHM = angle ABM; <br />also angle CEM = angle ABE (ACEM on the same circle O), <br />so CE//MH<br /><br />Let GN and OF cross at P;<br />OA=OB, so curve OA=curve OB;<br />curve AG= curve GD; curve BN=curve NF; so angle(OA)+angle(OB)+angel(AG)+angle(GD)+angle(DMHF)+angle(NF)+angle(BN)=180; <br />so angle NPF = angle(NF)+angle(OA)+angle(AG) = 90-1/2 angle(DMHF) =90- 1/2angleFOD = angle CEO.<br /> So CE//GN;<br />So CE//GN//MH<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-75958725771960328352019-07-13T19:28:03.140-07:002019-07-13T19:28:03.140-07:00https://photos.app.goo.gl/VcpDVKKDqwxshkUg9 Defin...https://photos.app.goo.gl/VcpDVKKDqwxshkUg9<br /><br />Define points I, J, K, L, P, S,T, X and Y as per attached sketch<br />Note that I, J are midpoints of OD and CD<br />L and K are midpoints of OF and EF<br />QXJI and QYKL are rectangles<br />We have OJ= ½(OC+OD)= 1/2OC + OI<br />And JI= OJ-OI= 1/2OC= QX<br />Similarly, KL= ½ OE=1/2OC= QY<br />Since QX=QY =&gt; Chord GH= Chord MN=&gt; Arc GH= Arc MN<br />And Arc GM=Arc NH =&gt; GN//MH<br />Let OS ⊥GN , we have quadrilaterals OSGJ and OSNK are cyclic <br />We have ∠ (COS)= ∠ (SGJ)= u=∠ (SNK)= ∠ (SOP)<br />So ∠ (COS)=u=∠ (SOE) =&gt; OS is the angle bisector of ∠ (TOP) and ∠ (COE)=&gt; CE//GN <br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-13622564726415043682019-07-10T13:18:18.849-07:002019-07-10T13:18:18.849-07:00See a simple solution on this drawing<a title="See the drawing" href="https://drive.google.com/file/d/1WW-4iz_Tw5QBk_z75IRIV94lq7cDay8p/view?usp=sharing" rel="nofollow"> See a simple solution on this drawing</a><br />rv.littlemanhttps://www.blogger.com/profile/15820037721044128612noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-40445788717072929422019-07-09T14:54:40.957-07:002019-07-09T14:54:40.957-07:00If: OC=r , OD=a =&gt; CM=MD=(a-r)/2 , OM=(a+r)/2 ...If:<br /><br />OC=r , OD=a =&gt; CM=MD=(a-r)/2 , OM=(a+r)/2<br /><br />Define:<br />ME=p<br /><br />So:<br />tg(MEC)=(a-r)/2p =&gt; 90-arctg((a-r)/2p)=MCE<br /><br />tg(MEO)=(a+r)/2p =&gt; 90-arctg((a+r)/2p)=MOE<br /><br />We know:<br />1.) OAC=OCA and AOC=alpha =&gt; OAC=90-alpha/2<br /><br />2.) Points A O E give a Tri., so AOE+OEA+OAE=180<br /><br />So:<br />MEO-MEC+AOC+OAC+MOE=180 =&gt; alpha=2*arctg(a-r)/2p=2*MEO<br /><br />_____<br /><br />90-alpha/2=MCE =&gt; OCA=MCE <br /><br />So AC and CE are on the line.<br /><br /><br />Ludwig Mercknoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-90447130730171623622019-07-08T09:44:52.542-07:002019-07-08T09:44:52.542-07:00Together with my friend rv.littleman we came up wi...Together with my friend rv.littleman we came up with this :<br /><br />Draw AE bisector of Ang(CAD). Ang(AEB) = Ang(EAC) + Ang(ECA) = 3α = Ang(EAB). So ABE is isoscele in B and AB = EB.<br />ADC and AEC have same base AC and same base angle pairs α and 3α. They are similar and therefore CD = AE which is equal to AB (given).<br />So AB = EB = AE, ABE is equilateral and x = 60°.Greghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-62404775437080812322019-07-07T05:48:56.371-07:002019-07-07T05:48:56.371-07:00please explain why CE=BD ? it does not seem so evi...please explain why CE=BD ?<br />it does not seem so evident<br />regards<br />DanielGewürtztraminer68https://www.blogger.com/profile/00329980114313015297noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-87384013660719554422019-07-04T13:45:07.976-07:002019-07-04T13:45:07.976-07:00Extend OQ to meet cirlce Q at F and form the diame...Extend OQ to meet cirlce Q at F and form the diameter OF. <br />Let m(EDF)=x =&gt; ECD and OAC are two similar isosceles triangles with angles (2x,90-x,90-x)<br />=&gt; AC/OC=CD/EC<br />=&gt;AC.EC=OC.CD<br />Since AE and OD are two chords of Q =&gt; A,C,E must be collinearSailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-41855445369354614842019-07-01T08:40:12.625-07:002019-07-01T08:40:12.625-07:00Define B’ symmetric of B by AD =&gt; AB=AB’ ang BA...Define B’ symmetric of B by AD =&gt; AB=AB’<br />ang BAD = a therefore ang BAB’ = 2a<br />Therefore ΔBAB’ is congruent to ΔBCD, and BB’=BD, BD=DB’<br />Therefore ΔBB’D is equilateral<br />Draw a line passing through B and perpendicular to AC meeting in E<br />AB=BC =&gt; BE is angle bisector of ang ABC<br />The drawing is symmetric, therefore BE is also angle bisector of ang DBB’<br />AD is angle bisector of ang BDB’ and CB’ is angle bisector of ang BB’D<br />By symmetry lines AD, CB’ and EB meet in point P<br />P is on angle bisector of ang DBB’, on angle bisector ofang BB’D and on angle bisector of ang BDB’ therefore P is the center of ΔBDB’<br />Therefore PD=PB=PB’<br />Consider now ΔAPB<br />Since ΔBB’D is equilateral, ang APB = 120<br />ang APB +ang PBA +ang BAP = 180<br />120+x/2+a=180 =&gt;x+2a=120 =&gt; x=120-2a<br />rv.littlemanhttps://www.blogger.com/profile/15820037721044128612noreply@blogger.com