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in Triangles by (44.3k points)

In △ ABC, side AB is produced to D such that BD = BC. If ∠ A = 70° and ∠ B = 60°, prove that

(i) AD > CD

(ii) AD > AC.

1 Answer

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In △ ABC it is given that ∠ A = 70o and ∠ B = 60o

Based on the sum property of the triangle

∠ A + ∠ B + ∠ C = 180o

To find ∠ C

∠ C = 180o – ∠ A – ∠ B

By substituting the values in the above equation

∠ C = 180o – 70o – 60o

∠ C = 180o – 130o

By subtraction

∠ C = 50o

Consider △ BCD

We know that ∠ CBD is the exterior angle of ∠ ABC

So we get

∠ CBD = ∠ DAC + ∠ ACB

By substituting the values in the above equation

∠ CBD = 70o + 50o

By addition

∠ CBD = 120o

It is given that BC = BD

So we can write it as

∠ BCD = ∠ BDC

Based on the sum property of the triangle

∠ BCD + ∠ BDC + ∠ CBD = 180o

So we get

∠ BCD + ∠ BDC = 180o – ∠ CBD

By substituting values in the above equation

∠ BCD + ∠ BDC = 180o – 120o

∠ BCD + ∠ BDC = 60o

It can be written as

2 ∠ BCD = 60o

By division

∠ BCD = ∠ BDC = 30o

In △ ACD

It is given that ∠ A = 70o and ∠ B = 60o

We can write it as

∠ ACD = ∠ ACB + ∠ BCD

By substituting the values we get

∠ ACD = 50o + 30o

By addition

∠ ACD = 80o

So we get to know that ∠ ACD is the greatest angle and the side opposite to it i.e. AD is the longest side.

Therefore, it is proved that AD > CD

We know that ∠ BDC is the smallest angle and the side opposite to it i.e. AC is the shortest side.

Therefore, it is proved that AD > AC.

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