By conditions of the problem, we obtain:

A + 2B = 2*pi rad, where A = < BAC and:
< ACB = < ABC = B .......(I)

Also, since Triangle AED is isosceles, it follows that:
< AED = A, while < BDE = < AED + < ADE = 2A

Again, < CED = 2*pi - A = 2B in terms of (i)

We now draw BP which bisects < BCE and meets DB at the point Q. We join PQ.

Thus, triangles QCP and QCB are congruent, and accordingly:

< QPC = < QBC = B

But, < DPC = < DPQ + < CPQ
Or, 2B = < DPQ + B, giving:
< DPQ = B

Thus, it follows that the triangles DPQ and CPQ are congruent, giving:

< QDP = < QCP

But, < QDP = < QCP = 2A, and:
< QCP = = B/2, so that:
2A = B/2
Or, B = 4A

Substituting  this into equation (i), we obtain:
9A = 2*pi rad, giving A = 2*pi/9

Consequently, the required measure of the angle A is 2*pi/9 radians. 

Edited on August 21, 2007, 11:15 am

Posted by K Sengupta on 2007-05-31 05:19:37