(In reply to

Answer by K Sengupta)

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**