tag:blogger.com,1999:blog-6933544261975483399.post4020395353392509560..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 317: Right triangle and Inscribed SquaresAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6933544261975483399.post-26379876227861066792021-01-09T05:35:53.202-08:002021-01-09T05:35:53.202-08:00Further problem - Prove that AC = (a+b)(a^2 + ab +...Further problem - Prove that AC = (a+b)(a^2 + ab + b^2)/abSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-72746187427738430682021-01-08T01:45:05.834-08:002021-01-08T01:45:05.834-08:00https://photos.app.goo.gl/rdkuinauWGpHAdUE8https://photos.app.goo.gl/rdkuinauWGpHAdUE8c.t.e.o.https://www.blogger.com/profile/16937400830387715195noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-41620809136649798202018-03-18T22:52:49.108-07:002018-03-18T22:52:49.108-07:00Geometry solution:
https://photos.app.goo.gl/Hx32K...Geometry solution:<br />https://photos.app.goo.gl/Hx32KiWgwVlZtzIM2<br />Let BC= a and AH= h ( see sketch)<br />1. Draw altitude AH= h<br />2. Extend BT such that BT= AH= h<br />3. Connect AT<br />4. Draw BQ//AT . Q is on AC<br />5. Draw QP//BC. P is on AB<br />6. Draw QR ⊥BC. Q is on BC<br />7. Draw PS⊥BC. S is on BC<br />Perform calculation we will get PQ=QR= a.h/(a+h)<br />And PQRS is a Square<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65893964028016828882018-03-17T06:40:03.995-07:002018-03-17T06:40:03.995-07:00May I ask a question?
ABC is a right angled Triang...May I ask a question?<br />ABC is a right angled Triangle with right angle at A. How to inscribe a square PQRS such that P. Q lie on the sides AB, AC respectively and both R, S lie on the hypotenuse BC ?.<br />Vijaya Prasad Nalluri (Pravin)<br />Rajahmundry, INDIA.Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-21087776821575494222009-11-23T10:05:24.346-08:002009-11-23T10:05:24.346-08:00tr JKE and tr EBF are similar ( an JEK = an BFE )
...tr JKE and tr EBF are similar ( an JEK = an BFE )<br /><br />=> EB/BF = a/x-a <br /><br />tr EBF and MFN are similar ( an BFE = an FNM )<br /><br />=> EB/BF = x-b/b<br /><br />from above => a/x-a = x-b/b <br /><br />=> x = a + bAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-60325889184392912672009-07-10T21:40:03.438-07:002009-07-10T21:40:03.438-07:00Denoting AH & PC by c & d resply. and noti...Denoting AH & PC by c & d resply. and noting the fact that all rt. angled triangles within Tr. ABC are similar to it, we can say that c/(c+a)= a/x while d/(d+b)= b/x. Further, c/a = AB/BC and d/b = BC/AB or c/(c+a)= AB/(AB+BC) and d/(d+b)= BC/(AB+BC). This gives us, a/x + b/x = c/(c+a) + d/(d+b)= AB/(AB+BC) + BC/(AB+BC) = 1. In other words, x = a + b QED.Tophttps://www.blogger.com/profile/11386384970463939250noreply@blogger.com