Question

If D is a point on the side BC of a triangle ABC such that AD bisects angle BAC. Then, show that BAgreaterthan BD.

Answer 1

Step-by-step explanation:

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Answer 2

BA > BD

Step-by-step explanation:

We are given that D is a point on the side BC of a triangle ABC such that AD bisects angle BAC.

And we have to show that BA > BD.

Firstly, in $\triangle ABC$, it is stated that AD bisects angle BAC which means that ;

$\angle$BAD = $\angle$CAD   {as AD bisects $\angle$A} ------------ [Equation 1]

Now, as we know that the exterior angle of a triangle is greater than the interior opposite angles of the triangle.

So, considering $\triangle$ADC,

$\angle$BDA > $\angle$CAD     {$\because$ exterior angle > interior opposite angle}

Also, $\angle$BDA > $\angle$BAD        {using equation 1}

Further, the sides opposite to the greater angle is also greater which means that;

BA > BD    {because the side opposite to $\angle$BDA is BA and the side

                        opposite to $\angle$BAD is BD}

Hence proved that BA > BD.