Question

In Fig. 10.131, prove that:
(i) CD + DA + AB + BC > 2AC
(ii) CD+DA +AB > BC

Answer 1

Step-by-step explanation:

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Answer 2

Hence Proved

$(i) CD + DA + AB + BC > 2AC$

$(ii) CD+DA +AB > BC$

Step-by-step explanation:

We need to prove:

$(i) CD + DA + AB + BC > 2AC$

$(ii) CD+DA +AB > BC$

Solution:

Solving for part (i).

In Δ ABC

Now we know that;

By Triangle Inequality property which sates that;

"The sum of two sides of the triangle should be greater than third side."

so we can say that;

$AB+BC>AC$ ⇒ Equation 1

In Δ ADC

By Triangle Inequality property

$DA+CD>AC$ ⇒ Equation 2

Adding equation 1 and equation 2 we get;

$DA+CD+AB+BC>AC+AC\\\\DA+CD+AB+BC>2AC$

Hence Proved

Solving for part (ii)

In Δ ABC

By Triangle Inequality property;

$AC+AB>BC$ ⇒ Equation 3

In Δ ADC

By Triangle Inequality property

$DA+CD>AC$

Adding both side by AB we get;

$DA+CD+AB>AC+AB$

Substituting the values from equation 3 we get;

$DA+CD+AB>BC$

Hence Proved.