Solution for the value of an angle of a triangle ABC

Question

Find value of angle m< DBC

Where $$BD=DC=AC$$ $$2(m\langle BAC)=14(m\langle ABD)=7(m\langle BCD)$$

I tried hard but im out of ideas now, I know the answer is 20 but I want to know how, thanks in advance.

Answer 1

Refer to the figure.

2x = 14y = 7z (given)

After dividing throughout by 14, we have x/7 = y/1 = z/2 = k for some non-zero k.

Then, x = 7k; y = k; and z = 2k [note:- all measures are in degrees]

Angle DBC = [1] = z = 2k

[2] = 180 – 2z = 180 – 4k

[3] = 180 – z – [1] – y – x = … = 180 – 12k

[4] = (180 – [3]) / 2 = … = 6k

[5] = [4] = 6k

[6] = x – [5] = … = k

Thus, [6] = k = y. This means AD = DB = DC = CA implying that ⊿DAC is equilateral.

∴ [4] = 6k = 60 => k = 10

∴ [1] = 2k = 20 degrees

Comment:- This is a tricky question.