Dragan

154º

48º

26º

106º

26º

m = 67.5º

n = 56.25º

In triangle ABC, angle bisectors AD and BE meet at a point I. Prove that angle DIB = 90 - angle BCA/2.

I'm purposely using a right-angled triangle because it gives me a better overall picture. 

Angle ADB = 180º - (90º + 30º) = 60º

Angle  DIB = 90º ∠BCA /2               ∠DIB = 90º - 30º/2 = 75º

Look at a triangle BID                       ∠DIB = 180º - (45º + 60º) = 75º  

( This is one way to prove it...This may not be the answer they're looking for.)

Important: The Median Theorem states that the medians of a triangle intersect at a point called the centroid that is two-thirds of the distance from the vertices to the midpoint of the opposite sides.

ST² - TN² = SN²                

SX / NX = 2 / 1

SX = 2/3 * SN  

I'm gonna attempt to answer Guest's question: What is the angle APM?

∠BON = ∠AOT = 60º                  ∠APT = 30º

∠BOT = ∠AON= 120º

Radius  TO = 1 

Diameter TN = 2

PT = TO / tan(30º)

∠NPT = tanarc( TN / PT ) ≈ 49.107º

∠AMP ≈ 19.107º

(1st question is for 5th graders and the second one for college students.) 

The area of the shaded region is    A = 2 [ 2 / cos(30º)]

My pleasure!

When you put a regular pentagon and a square next to each other, their sides create an angle of 162º. 

The internal angle of a pentagon is 108º  and of a square 90º.        360º - ( 108º + 90º ) = 162º

I'll let you find that n-gon.  

check your numbers; something's not right

There are two different triangles: ΔABC  and  ΔADE   These triangles share the same angle,  ∠BAC  or  ∠DAE

1)       Use the Law of cosines to calculate an angle BAC of a triangle ABC.

2)       Use the Law of cosines to find the length of DE.