Given that in ΔABC, side AB is produced to D so that BD = BC and ∠B = 60°, ∠A = 70°
We have to prove that,
(i) AD > CD
And, (ii) AD > AC
First join C and D
Now,
In ΔABC,
∠A + ∠B + ∠C = 180°(Sum of all angles of triangle)
∠C = 180° – 70° – 60°
= 50°
∠C = 50°
∠ACB = 50°(i)
And also in ΔBDC,
∠DBC = 180° - ∠ABC (Therefore, ∠ABD is straight angle)
= 180° – 60°
= 120°
BD = BC (Given)
∠BCD = ∠BDC (Therefore, angle opposite to equal sides are equal)
Now,
∠DBC + ∠BCD + ∠BDC = 180°(Sum of all sides of triangle)
120° + ∠BCD + ∠BCD = 180°
2∠BCD = 180° – 120°
2∠BCD = 60°
∠BCD = 30°
Therefore, ∠BCD = ∠BDC = 30°(ii)
Now, consider ΔADC,
∠BAC = ∠DAC = 70°(Given)
∠BDC = ∠ADC = 30°[From (ii)]
∠ACD = ∠ACB + ∠BCD
= 50° + 30°[From (i) and (ii)]
= 80°
Now,
∠ADC < ∠DAC < ∠ACD
AC < DC < AD (Therefore, side opposite to greater angle is longer and smaller angle is smaller)
AD > CD
And,
AD > AC
Hence, proved
We have,
∠ACD > ∠DAC
And,
∠ACD > ∠ADC
AD > DC
And,
AD > AC (Therefore, side opposite to greater angle is longer and smaller angle is smaller)
