Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Saturday, February 23, 2019
Geometry Problem 1415: Right Triangle, Altitude, Incircle, Excircle, Tangency Points, Isosceles Triangle
Labels:
altitude,
excircle,
geometry problem,
incircle,
isosceles,
right angle,
right triangle,
tangency point
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∆JBF isosceles => ∆JGB isosceles
ReplyDelete< BFJ = C/2 from Problem 1410 and < ABJ = C so < BJF = C/2 and hence BJ = BF = BG
ReplyDeleteSumith Peiris
Moratuwa
Sri Lanka
https://photos.app.goo.gl/yqpR9Fb7fB8CnjAw6
ReplyDeleteDefine points L ,K and angle u per sketch
Per the results of previous problems we have ICE is a isosceles right triangle and DJ ⊥IC
We have GC= p-AC= BD ( p= half perimeter of trị ABC)
Triangle EGC congruent to FBD( case SAS)
So ∠ (BFD)= ∠ (GEC)=u
We also have ∠ (HCI)=u=∠ (BJD)=∠ (BFJ)=> JBF is isosceles
But BG=BF= BJ => JBG is isosceles