Question

In the given figure,prove that AB>DC

Answer 1

◆Ekansh Nimbalkar◆
hello friend here is your required answer

Answer 2

Given -- ABC is a triangle.
In ΔABD , AB = AD
∠ABD = 70°
∠ACD = 40°

To prove - AB > DC

Proof -- Since AB = AD
∠ABD = ∠ADB     (By isosceles triangle property)
∠ABD = ∠ADB = 70° ------(i)

Now, ∠ADB  + ∠ADC = 180° (By linear pair property)
70° + ∠ADC = 180°  (From (i))
∠ADC = 110°

Now in ΔADC,
∠ADC + ∠ACD + ∠DAC = 180°  (Angle sum property of triangle)
110 + 40 + ∠DAC = 180
∠DAC = 180 - 150
∠DAC = 30°

40° > 30°
∠ACD > ∠DAC
AD > DC  (The side opposite to the greater angle is greater)

But we know AD = AB
So , AB > DC

Hence Proved!

Hope This Helps You!