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

Level: Mathematics Education, High School, Honors Geometry, College.

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## Wednesday, April 5, 2017

### Geometry Problem 1326: Triangle, Cevian, Incenters, Sum of Angles, 270 Degrees

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Problem 1326

ReplyDeleteIs <AFB+<BFC=(90+<BDA/2)+(90+<BDC/2)=180+(<BDA/2+<BDC/2)=180+90=270.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

https://goo.gl/photos/uQbYvyPmU8Kd2Bad6

ReplyDeleteSince E is the incenter of triangle ABD so we have

∠ (BEx)= ½ ∠ (D1)+1/2. ∠ (B1)

∠ (AEx)= ½. ∠ (D1)+ ½.∠ (A)

Add above side by side we have

∠ (E )= ∠ (D1)+ ½.∠ (A)+ ½.∠ (B1)

Replace ½.∠ (A)+ ½.∠ (B1)= 90- ½.∠ (D1) we will get ∠ (E )= 90+ ½.∠ (D1)

Similarly with incenter F we have ∠ (F)= 90+1/2. ∠ (D2)

D1 supplement to D2 so ∠ (E )+ ∠ (F)= 180+ ½.∠ (D1)+ 1/2. ∠ (D2) = 270

ANGLE AEB= 90+ 1/2 ANGLE ADB

ReplyDeleteANGLE CFB= 90+ 1/2 ANGLE CDB

ANGLE ADB+ ANGLE CDB = 180 DEG

THEREFORE ANGLE AEB+ANGLE CFB = 270DEG

Easy, can do without pen and paper.

ReplyDeleteIn ∆s AEB and BFC the sum of the angles,

<A/2 + <B/2 + <C/2 + <AEB + < BFC = 360

from which the result follows.

Sumith Peiris

Moratuwa

Sri Lanka

3 angles of the of the concave pentagon AEBFC are A/2, B/2 and C/2 which sum up to 90

ReplyDeleteHence (360 - <AEB) + (360 - < BFC) + 90 = 540 and the result follows

Sumith Peiris

Moratuwa

Sri Lanka

From this equation we may infer that < AEB and < BFC are both necessarily obtuse.

ReplyDeleteE+F=EBO+EOB+FBO+FOB=(EBO+FBO)+360-AEC=B/2+180+A/2+C/2=270

ReplyDelete