In Δ ABC, side AB is produced to D so that BD = BC.
∠ B = 60°, and ∠ A = 70°
To prove: (i) AD > CD (ii) AD > AC
Construction: Join C and D
We know, sum of angles in a triangle = 180°
∠ A + ∠ B + ∠ C = 180°
70° + 60° + ∠ C = 180°
∠ C = 180° – (130°) = 50°
∠ C = 50°
∠ ACB = 50° ……(i)
And also in Δ BDC
∠ DBC =180° – ∠ ABC = 180 – 60° = 120° [∠DBA is a straight line] and BD = BC [given]
∠ BCD = ∠ BDC [Angles opposite to equal sides are equal]
Sum of angles in a triangle = 180°
∠ DBC + ∠ BCD + ∠ BDC = 180°
120° + ∠ BCD + ∠ BCD = 180°
120° + 2∠ BCD = 180°
2∠ BCD = 180° – 120° = 60°
∠ BCD = 30°
∠ BCD = ∠ BDC = 30° ….(ii)
Now, consider Δ ADC.
∠ DAC = 70° [given]
∠ ADC = 30° [From (ii)]
∠ ACD = ∠ ACB + ∠ BCD = 50° + 30° = 80° [From (i) and (ii)]
Now, ∠ ADC < ∠ DAC < ∠ ACD
AC < DC < AD [Side opposite to greater angle is longer and smaller angle is smaller]
AD > CD and AD > AC
Hence proved.
