tag:blogger.com,1999:blog-6933544261975483399.post2282266921620117638..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1489: Right Triangle, Angle Bisectors, Perpendicular, MeasurementAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6933544261975483399.post-49374660730838045362021-09-27T23:08:30.703-07:002021-09-27T23:08:30.703-07:00ITS a good geometrITS a good <a href="https://gogeometry.blogspot.com/2020/12/geometry-problem-1489-right-triangle.html#comment-form" rel="nofollow">geometr</a>jkhatrihttps://www.blogger.com/profile/09562840913640450433noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-11103966225612540432020-12-04T09:37:50.432-08:002020-12-04T09:37:50.432-08:00https://photos.app.goo.gl/TZto12GVkGKjP58p6https://photos.app.goo.gl/TZto12GVkGKjP58p6c.t.e.o.https://www.blogger.com/profile/16937400830387715195noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-19877338601871876822020-12-04T04:39:53.432-08:002020-12-04T04:39:53.432-08:00Draw perpendicular from D to meet AC at P, BC at Q...Draw perpendicular from D to meet AC at P, BC at Q. DP=DQ=radius of incircle = 2 <br />Extend AD to meet BC at O. Since AO is Angle bisector BO/OC=3/5=> BO=3,OC=5<br />As PED, BAO,PAD are similar<br />=> PE=1, AP=4 => EC =5 and E is mid-point of AC and since EF||AB=>FC=4<br />Hence OF=1. Since BQ=2=> QO=1 <br />So DQF is isosceles right triangle where DQ=QF=2=> DF=2.Sqrt(2)Sailendra Thttps://www.blogger.com/profile/12056621729673423024noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-73828680293343548382020-12-04T04:37:39.613-08:002020-12-04T04:37:39.613-08:00Extend AD to P, P on BC. DPFE cyclic => Tr BDF ...Extend AD to P, P on BC. DPFE cyclic => Tr BDF rightisosceles<br />=> DF=DB=2V2c.t.e.o.https://www.blogger.com/profile/16937400830387715195noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-50719152324526554732020-12-04T02:19:56.390-08:002020-12-04T02:19:56.390-08:00Inradius of Tr. AB, r = S/s = (48/2)/12 = 2
If U ...Inradius of Tr. AB, r = S/s = (48/2)/12 = 2<br /><br />If U is the tangent point of the incircle on AC, AU = s - a = 12 - 8 = 4 and so <br />From similar Tr.s AUD & DUE, UE = 1, so E is the mid point of AC & F is the mid point of BC (mid point theorem)<br /><br />Hence BF = 4<br /><br />If V is the point of tangency on BC of the incircle, BV = s-b = 12-10 = 2<br />Hence VF = 4-2 = 2<br /><br />So Tr.s BVD & FVD are congruent SAS being also right isosceles and DF = BD = 2.sqrt2<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-75973265486083254432020-12-03T20:50:49.678-08:002020-12-03T20:50:49.678-08:00https://photos.app.goo.gl/wzBcnmF6f4N9989D6
Draw ...https://photos.app.goo.gl/wzBcnmF6f4N9989D6<br /><br />Draw incircle D with radius= r<br />Define position of M, N and P ( see sketch)<br />Let ∠ (A)=2u and ∠ (C)=2v<br />We have ∠ (DEF)=180-∠ (DEA)- ∠ (FEC)= u + 2v<br />Replace 2v= 90- 2u we get ∠ (DEF)=90-u=∠ (DEA)<br />Triangle DME congruent to DPE ( case HA) => EF tangent to incircle D<br />So DPFN is a square and DF= r. sqrt(2)<br />But r=2 ( see problem 1487) so DF=2.sqrt(2)<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com