Geometry Problem 1488. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.
Details: Click on the figure below.
Wednesday, December 2, 2020
Geometry Problem 1488: Right Triangle, Altitude, Angle Bisectors, Measurement
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FEC=45, C=40 => EFC=85 => EFB=95
ReplyDeletehttps://photos.app.goo.gl/cA2eVoGjetr1yqXc7
ReplyDeletefrom the result of problem 1487 we have CM⊥BE and ∠ (ADM)= 45 and triangle ECB is isosceles
in triangle BDC, external angle ∠BDM)= ∠ (DCB)+ ∠ (BDC)= 65
but ∠ (EDM)= ∠ (BDM)= 65 ( symmetric in isosceles triangle EBC)
we have ∠ (ADC)= 90+∠ (B/2)= 145 => ∠ (ADM)= 45 => ∠ (ADC)= 65-45=20
in right triangle AND, ∠ (ADN)=90-25= 65 => ∠ (EDN)=65-20=45=∠ (DEC)
in triangle EFC, external angle ∠ (BFE)= ∠(FEC)+ ∠ (FCE)= 85
Reference my proof of Problem 1487, ABDE is concyclic, so < BDF = BAE = 50
ReplyDeleteand so < BFD = 180 - 45 - 50 = 85
Sumith Peiris
Moratuwa
Sri Lanka