Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

## 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 trá»‹ 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