## Saturday, February 23, 2019

### Geometry Problem 1416: Right Triangle, Altitude, Incircle, Excircle, Tangency Points, 45 Degree Angle

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

Details: Click on the figure below. 1. Draw FT, T on AC, Extend JG to P, P on FT => ∆GPT right isosceles,
=> ang PJF = JFT = 45°

2. From my proof of Problem 1415, BF = BD = BJ so B is the centre of Tr. HFG

So Alpha = < GBF /2 = 90/2 =45

Sumith Peiris
Moratuwa
Sri Lanka

3. https://photos.app.goo.gl/syrJ1dYNJgpap7oi9

See sketch for position of points L, K and angle u
per the results of previous problems we have JBG is isosceles and IC ⊥JD and ICE is isosceles right triangle.
Draw angle bisector of angle HBD, this line meets IN at L
In right triangle ABC we have ∠ (BAH) = 2u= ∠ (HBC)
So ∠ (BAI)= ∠ (LBN)= u => ∠ (BLN)=90=> AL//JG
So ∠ (IKJ)= ∠ (KIL)= 45 and ∠ (LJK)= 45

4. Second way
Draw second diagonal of trapezoid, let be P intersection
∆PJG right isosceles