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

Labels:
45 degrees,
altitude,
angle,
excircle,
geometry problem,
incircle,
right triangle,
tangency point

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Draw FT, T on AC, Extend JG to P, P on FT => ∆GPT right isosceles,

ReplyDelete=> ang PJF = JFT = 45°

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

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

Sumith Peiris

Moratuwa

Sri Lanka

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

ReplyDeleteSee 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

Second way

ReplyDeleteDraw second diagonal of trapezoid, let be P intersection

∆PJG right isosceles