## Saturday, October 10, 2015

### Geometry Problem 1152: Triangle, Quadrilateral, Angle, 10, 20, 30 Degrees

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

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.
Click the diagram below for more details. 1. http://s3.postimg.org/ewu8vbicj/pro_1152.png

Draw Av//CD
We have ∠ (DAv)=180-50=130
And ∠ (DAv)= ∠ (DAC)+ ∠ (CAx)= ∠ (BAD)-x+∠ (ACD)=145-x
So x=15

2. Would it be correct to say that since /_ABC=155°=Larger /_ADC)/2=310°)/2 and /_ADB=2*/_BCA, a circle drawn with centre at D & radius = DA will pass through A, B & C? This would imply /_x=30°/2=15°.

3. We observe that /_BCA=10° while /_ADB =20° & simultaneously /_ADC=155° =(Larger /_CDA)/2.
Hence would it be correct to say that the circumcircle of Tr. ABC will have point D as its centre and DA(=DB=DC) as its radius? If that be correct then /_BAC = x = /_BDC/2 =30/2=15°.

1. It is but it needs to be proved Ajit

I'm sure u know the proof

2. Please elaborate, Sumith, and if possible please write to me at: ajitathle@gmail.com.

4. Triangle DAC: <(CAD) + <(ACD) = 180 - 50 = 130. And <(BAD) +<(ACD)= 145.
So x=<(BAD) + <(ACD) - <(CAD) - <(ACD) =145 -130= 15.

5. See triangle ABC and triangle ABD, AB is common, and ang.ADB=2*ang.ACB.
So, A, B, and C are cocyclic (circum angle/central angle).

---> x=ang..BDC/2 = 15 deg.

6. Please note all above solutions never use given data m(ACB)= 10 degrees. in fact m(ACB) can have value not equal 10 degrees.

Peter Tran