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 1414: Right Triangle, Altitude, Incircle, Excircle, Tangency Points, Isosceles Triangle

Labels:
altitude,
excircle,
geometry problem,
incircle,
isosceles,
right triangle,
tangency point

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Join E to G, draw ET ꓕ AC, => ∆DBJ~∆GET

ReplyDeleteLet the bisector of < JBF cut DF at U.

ReplyDeleteFrom my proof of Problem 1412, < DGF = C/2, hence < GDF = < BFD - < DGF = 45 - C/2 = A/2 = < JBU since < JBF is obviously = to A in right Tr. ABC

Hence BDJU is concyclic and < BJU = < BDF = 45

So Tr.s BJU and BFU are congruent ASA and therefore

BJ = BF

Sumith Peiris

Moratuwa

Sri Lanka

https://photos.app.goo.gl/UvMged3MaAHNfXMy6

ReplyDeleteDefine point M and angle u per sketch

Per the results of previous problems we have D,G,M are colinear

In right triangle DBG, angle BDG complement to u

In right triangle JHM, angle ∠ (HJM)= ∠ (BJD) complement to u

So ∠ (BDJ)= ∠ (BJD) and BD=BJ=BF -> JBF is isosceles