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

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

Click the figure below to see the complete problem 1018.

## Saturday, May 24, 2014

### Geometry Problem 1018: Right Triangle, Circles, Angle

Labels:
angle,
circle,
right triangle

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∠DBE＝180。－∠BDE－∠BED

ReplyDelete＝180。－1/2（180。－∠A）－1/2（180。－∠C）

＝180。－90。＋1/2∠A－90。＋1/2∠C

＝1/2（∠A＋∠C）

＝1/2（180。－∠ABC）

＝1/2（180。－90。）

＝45。

Consider angle CBD as X ---> angle BAC = 2X (tangents rule)

ReplyDeleteConsider angle ABE as Y ---> angle BCA = 2Y (tangents rule)

From here we can see in triangle ABC that 2X + 2Y = 90 --> X+Y = 45

In triangle BDE we get :

angle BDE = 2Y + X

angle BED = 2X + Y

angle BDE + angle BED = 3Y + 3X = 135 THUS DBE = 45

<ABE=<BCE/2, and <CBD=<BAD/2, so <EBD=90-<ABE-<CBD=90-(<BCE/2+<BAD/2)=90-90/2=45

ReplyDelete3 Angles of Tr. BED =< ABC + 2.<EBD. Hence < EBD = 90/2= 45

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka