## Wednesday, April 5, 2017

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

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below. 1. Problem 1326
Is <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

2. Since 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

3. ANGLE AEB= 90+ 1/2 ANGLE ADB
ANGLE CFB= 90+ 1/2 ANGLE CDB
ANGLE ADB+ ANGLE CDB = 180 DEG

THEREFORE ANGLE AEB+ANGLE CFB = 270DEG

4. Easy, can do without pen and paper.

In ∆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

5. 3 angles of the of the concave pentagon AEBFC are A/2, B/2 and C/2 which sum up to 90

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

Sumith Peiris
Moratuwa
Sri Lanka

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

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