In Δ ABC, side AB is produced to D so that BD=BC. If ∠B=60° and ∠A=70°, prove that:
(i) AD > CD (ii) AD > AC
Answers
Answer:
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Given: In Δ ABC, side AB is produced to D so that BD = BC. If ∠B = 60° and ∠A = 70°.
To Prove : (i) AD > CD (ii) AD > AC
Construction: Join C to D.
Proof :
In ∆ ABC, we have :
We know that, the sum of all angles of triangle is 180° :
∠ A + ∠ B + ∠ C = 180°
70° + 60° + ∠ C = 180°
∠ C = 180° – (130°) = 50°
∠ C = 50°
∠ ACB = 50° ………...…(1)
∠ DBC + ∠ ABC = 180°
[∠DBA is a straight line]
∠ DBC = 180° – ∠ ABC
∠ DBC = 180 – 60°
∠ DBC = 120°
But,
BD = BC (Given)
∠ BCD = ∠ BDC ……………(2)
[Angles opposite to equal sides are equal]
In ∆ BDC , we have :
We know that, the sum of all angles of triangle is 180°.
∠ DBC + ∠ BCD + ∠ BDC = 180°
120° + ∠ BCD + ∠ BCD = 180°
[From eq 2]
120° + 2∠ BCD = 180°
2∠ BCD = 180° – 120°
2∠ BCD = 60°
∠ BCD = 60°/2
∠ BCD = 30°
∠ BCD = ∠ BDC = 30° ……………(3)
Now, in Δ ADC,
∠ DAC = 70° [given]
∠ ADC = 30° [From eq 3]
∠ ACD = ∠ ACB + ∠ BCD
∠ ACD = 50° + 30°
[From eq (1) and 3]
∠ ACD = 80°
Now, ∠ ADC < ∠ DAC < ∠ ACD
AC < DC < AD
[Side opposite to greater angle is larger and smaller angle is smaller]
AD > CD and AD > AC
Hence proved.
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