Question

In a triangle abc side ab is produced to d such that bd =bc if angle a =70 and angle b=60 prove that ad>cdand ad>ac

Answer 1

Answer:

Step-by-step explanation:

Given that in

, side AB is produced to D so that BD = BC and ∠B = 60o, ∠A = 70o

We have to prove that,

(i) AD > CD

And, (ii) AD > AC

First join C and D

Now,

In

,

∠A + ∠B + ∠C = 180o (Sum of all angles of triangle)

∠C = 180o – 70o – 60o

= 50o

∠C = 50o

∠ACB = 50o (i)

And also in

,

∠DBC = 180o - ∠ABC (Therefore, ∠ABD is straight angle)

= 180o – 60o

= 120o

BD = BC (Given)

∠BCD = ∠BDC (Therefore, angle opposite to equal sides are equal)

Now,

∠DBC + ∠BCD + ∠BDC = 180o (Sum of all sides of triangle)

120o + ∠BCD + ∠BCD = 180o

2∠BCD = 180o – 120o

2∠BCD = 60o

∠BCD = 30o

Therefore, ∠BCD = ∠BDC = 30o (ii)

Now, consider

,

∠BAC = ∠DAC = 70o (Given)

∠BDC = ∠ADC = 30o [From (ii)]

∠ACD = ∠ACB + ∠BCD

= 50o + 30o [From (i) and (ii)]

= 80o

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)