Geometry Problem 1504. Post your solution in the comment box below.

Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

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Tr. CEB is isoceles with angles 15,15,150

ReplyDeleteTr.s CEI & BEF are congruent SAS

If < CEI = < BEF = p then considering the angles in Tr. BEC, 15+15+p + p + 90-p = 180

So p = 60 and < IEF = 30 and so < GJH = 150

Sumith Peiris

Moratuwa

Sri Lanka

2 interesting results

Delete1 - FC is perpendicular to AC

2 - Tr. BIF and Tr. JFH have angles 15,30 and 135

A simpler solution

ReplyDeleteSince < BEC = 150 and < FEC = 90, < BEF = 60 so < EFI = 30

Hence considering cyclic quadrilateral EIJF, < GJH = 150

Sumith Peiris

Moratuwa

Sri Lanka

https://photos.app.goo.gl/poMiXSFHd2rRzTxS6

ReplyDeleteConstruct equilateral triangle CBP as shown

Note that triangle CBP is the translation of triangle DAE with EP=DC and DE=CP

Since EP is an angle bisector of ∠(CPB) > ∠(CPE)=30

Since EP=CP -> CEP is isoceles - > ∠(CEP)=∠(ECP)= 75 and ∠(ECB)= 15

So ∠(CEB)= ∠(GJH)= 150