## Saturday, February 23, 2019

### Geometry Problem 1414: Right Triangle, Altitude, Incircle, Excircle, Tangency Points, Isosceles Triangle

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

#### 3 comments:

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

2. Let the bisector of < JBF cut DF at U.

From 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

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