Question to Find the Unknown Value of a Angles

Topic : Similar Triangles

Question :

If two triangles with vertex A and ECD are similar to each other. If ∟A + ∟ECD = 80º , then find ∟B + ∟D ?

Solution :

∟∆BAC ~ ∆ECD
So , ∟A = ∟EDC
∟B = ∟DEC
∟ACB = ∟D

AB = AC (as ∟BAC is an isosceles triangle)
DE = DC(as ∟ECD is an isosceles triangle)
So, AB = ∟B = ACB
And ∟DEC = ∟ECD

Now ∟A + ∟ECD = 80º (given)
So, ∟A = ∟ECD = 80º/2 = 40º (as ∟A = ∟ECD)

In ∆ABC, ∟A + ∟B + ∟ACB = 180º
40º + ∟B + ∟B = 180º (as ∟A = 40º, ∟B = ∟ACB)
2 ∟B = 180º - 40º
∟B = 140º / 2
∟B =70º

Similarly, in ∆ECD, ∟D =70º
So, ∟B + ∟D = 70º + 70º =140º