Question

question number 18 please solve it​

Answer 1

Answer:

As already marked in your diagram, let x = ∠A, y = ∠B and z = ∠EDC.

Since AED is isosceles:

∠DEA = ∠A = x  (already marked in the diagram)

Since CDE is isosceles:

∠ECD = ∠EDC = z  (already marked in the diagram)

Since BEC is isosceles:

∠BEC = ∠B = y  (already marked in the diagram).

Since ∠EDC is external angle of triangle AED:

z = 2x      ... (1)

Since the angles on a straight line add up to 180° and so do the angles in a triangle:

x + y = ∠DEA + ∠BEC = 180° - ∠CED = 2z      ... (2)

Putting (1) and (2) together gives:

x + y = 2z = 2(2x) = 4x  =>  y = 3x

=> x : y = 1 : 3

=> ∠A : ∠B = 1 : 3

From (2), we have

x + y = 2z

=> x + y = z + z

=> x + y = z + 2x    [ from (1) ]

=> y = x + z      ... (3)

Since AB = AC, triangle ABC is isosceles so ∠ACB = ∠ABC = ∠B = y.

So...

∠BCE = ∠ACB - ∠ADE = y - z = x = ∠AED, as required.