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

Labels:
altitude,
angle,
bisector,
geometry problem,
incenter,
measurement,
right triangle

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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