Question

In the adjoining figure AB > AC and D is any point on BC. Show that AB > AD​

Answer 1

Answer :

AB > AD

Step-by-step explanation:

In ∆ABC is the angle opposite to side AB ∠ACB.

The angle opposite to side AC is ∠ABC.

Since,

AB > AC

Now,

⇒ ∠ACB >∠ABC        ......Equation(1)

Now,

Similarly here,

In ∆ADC,

⇒ ∠ADB =∠ACD+∠ DAC

⇒ ∠ADB= ∠ACB +∠DAC

Implies,

⇒ ∠ADB > ∠ACB       ........Equation(2)

Now, We have :

⇒ ∠ADB > ∠ACB       ........Equation(2)

⇒ ∠ACB >∠ABC        ......Equation(1)

So,

From Eq(1) and(2).

⇒ ∠ADB > ∠ABC.

Now in ∆ABD ∠ADB >∠ABD

Hence,

AB >AD.

$\rule{300}{1.5}$

★ Points to remember :

• Greater angle has greater side opposite to it.
• Angles opposite to bigger side is bigger and in simliar way side opposite to bigger angle is larger in the triangle.
• An exterior angle of a triangle is equal to the sum of the opposite interior angles.

Answer 2

Answer :

AB > AD

Step-by-step explanation:

In ∆ABC is the angle opposite to side AB ∠ACB.

The angle opposite to side AC is ∠ABC.

Since,

AB > AC

Now,

⇒ ∠ACB >∠ABC ......Equation(1)

Now,

Similarly here,

In ∆ADC,

⇒ ∠ADB =∠ACD+∠ DAC

⇒ ∠ADB= ∠ACB +∠DAC

Implies,

⇒ ∠ADB > ∠ACB ........Equation(2)

Now, We have :

⇒ ∠ADB > ∠ACB ........Equation(2)

⇒ ∠ACB >∠ABC ......Equation(1)

So,

From Eq(1) and(2).

⇒ ∠ADB > ∠ABC.

Now in ∆ABD ∠ADB >∠ABD

Hence,

AB >AD.

★ Points to remember :

Greater angle has greater side opposite to it.

Angles opposite to bigger side is bigger and in simliar way side opposite to bigger angle is larger in the triangle.

An exterior angle of a triangle is equal to the sum of the opposite interior angles.