Question

$\small{Hlo frndz}$
D is a point on side BC of triangle ABC such that AD=AC.Show that AB>AD.​

Answer 1

Given,

AD = AC

We know,

Angles opposite to equal sides are equal,

We get,

Angle (ACD) = Angle (ADC)

Again, we know,

Angle (ADC) is an exterior angle for triangle (ABD),

Exterior angle = sum of opposite interior angles

So,

Angle(ADC) = Angle(ABD) + Angle(BAD)... (1)

From (1) we can already say,

Angle (ADC) > Angle (ABD)

Then,

AB > AC

Because we know,

Side opposite to greater angle is greater,

Then,

AB > AD

Because we already know AD = AC,

Hence proved.

Answer 2

Answer:

Step-by-step explanation:

Given :

In ∆ABC ,

D i s a point on BC such that AD=AC

To show/prove :

AB > AD

Proof :

In ∆ ABC,

AD= AC

=> ∠ACD = ∠ ADC ..................... (1)

( °.° angles opposite to equal sides are equal )

Now,

for ∆ ABD,

∠ ADC is an exterior angle.

But,

We know that,

exterior angle equals to the sum of opposite interior angles

So,

∠ ADC = ∠ ABD + ∠ BAD

=> ∠ ACD = ∠ ABD + ∠ BAD ..............(2)

( °.° ∠ ADC = ∠ ACD from ${eq}^{n}$ 1 )

Now, In ∆ ABC

∠ ACD = ∠ ABD + ∠ BAD

( from ${eq}^{n}$ 2 )

=> ∠ ACD > ∠ ABD

( °.° ∠ ACD is sum of ∠ ABD and ∠ BAD )

=> ∠ ACB > ∠ ABC

( °.° points B and D both lies on same line )

=> AB > AC

( °.° side opposite to the greater angle is greater )

=> AB > AD

( °.° AC = AD )

Hence,

Proved