tag:blogger.com,1999:blog-6933544261975483399.post6079729409323534121..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1282 Squares, Common Vertex, Midpoint, Perpendicular, 90 DegreesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6933544261975483399.post-10725877217261582042018-03-21T17:31:44.466-07:002018-03-21T17:31:44.466-07:00see link below for the sketch of the problem 1282
...see link below for the sketch of the problem 1282<br /><br />https://photos.app.goo.gl/pN56UvkWo7gKq2n23<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-56184815029506871722018-03-21T02:02:23.321-07:002018-03-21T02:02:23.321-07:00I wanted to open the sketch but I couldn't so ...I wanted to open the sketch but I couldn't so now I don't understand the solution. Could someone please help me. I have to solve this for my mathclassAnonymoushttps://www.blogger.com/profile/02098133615673398529noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-74306866392152689682016-12-29T13:48:06.737-08:002016-12-29T13:48:06.737-08:00BY CONGRUENCE D1D2=B2B1, BB1=DD1 AND D1D2 ┴ B2B1 ,...BY CONGRUENCE D1D2=B2B1, BB1=DD1 AND D1D2 ┴ B2B1 , BB1 ┴ DD1 <br />Let X be the intersection of D1D2 AND B2B1,Y be the intersection of BB1 AND DD1 <br />THEN ( C, Y, B1, A1, X, D1 ):CONCYCLIC <br />∠(CYD1)=45=∠(XA1D1) => ∠(XD1A1)=∠(YB1C). SO,∠(BB1B2)=90 ISW ∠(DD1D2)=90 <br /><br /> Triangle BB1B2 ≡ DD1D2 ( SAS ) <br /> M : triangle BB1B2 with the circumcenter <br /> So BB2=DD2=2*MB1<br /><br /><br />Let W be the intersection of MB1 AND DD2<br /> ∠YDD2= ∠B1BB2= ∠MB1B THEN (D.Y.W.B1) CONCYCLIC => ∠DYB1=∠DWB1=90Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-5277713606672739272016-11-06T08:38:52.159-08:002016-11-06T08:38:52.159-08:00Thanks Peter
Another conclusion in your step 2:
DD...Thanks Peter<br />Another conclusion in your step 2:<br />DD2 = 2(B1M)Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-67904870583420007902016-11-06T00:34:33.410-07:002016-11-06T00:34:33.410-07:00https://goo.gl/photos/QQWPcN1k1LiZB8Jt8
Let N, Q,...https://goo.gl/photos/QQWPcN1k1LiZB8Jt8<br /><br />Let N, Q, E, F, P and R are the projection of B, D, M, B1, B2 and D2 over CA1 ( see sketch)<br />See sketch for positions of K, G , L and H.<br />Let u= D2R=A1P<br />v=B2P=A1R<br />v1=BN=CQ<br />u1=CN=DQ<br />w=FB1=FA1=FC=FD1<br />1. Since M is the midpoint of BB2=> E is the midpoint of NP<br />By observation we have EF=MG=NF-1/2.NP= ½(u1-u)<br />ME=1/2(v-v1)=GF<br />GB1= w+1/2(v-v1)<br />D2K=QR= v+2w-v1= 2.GB1<br />DK=DQ-QK=u1-u=2.MG<br />So MG/DK=GB1/D2K= ½<br />2. Triangle MGB1 similar to DKD2 ( case SAS)<br />So ∠ (MB1G)= ∠ (DD2K)=> D2,L, H, B1 are cocyclic <br />And ∠ (B1HD2)= ∠ (B1LD2)= 90 degrees<br />So B1M ⊥DD2<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com