Choose whether the given statement is true or false.

In a scalene triangle ABC, D is a point on the side AB such that CD2=AD.DB , if $sinAsinB = \frac{C}{2}$ then prove that CD is internal bisector of ∠C.

Question

Choose whether the given statement is true or false. In a scalene triangle ABC, D is a point on the side AB such that CD2=AD.DB , if sinAsinB=C2   then prove that CD is internal bisector of ∠C.

Accepted Answer

We draw a scalene triangle ABC having point D on side AB.

https://www.vedantu.com/question-sets/b6cdc4af-6191-477a-8c0b-82b361b830c78865827341593868215.png

Here ∠C is divided into two angles by the line CD i.e. ∠ACD and ∠BCD.
Let ∠ACD=x then we can write ∠BCD=C−x
Now we apply the sine rule in both triangles formed.
In △ACD,

\Rightarrow \frac{AD}{sinx} = \frac{CD}{sinA} . . . . . . . . . . . . . . . \dots (1)

In △BCD,

\Rightarrow \frac{BD}{\sin (C - x)} = \frac{CD}{sinB}

................… (2)
Now we multiply both the equations (1) and (2)

\Rightarrow \frac{AD}{sinx} \times \frac{BD}{\sin (C - x)} = \frac{CD}{sinB} \times \frac{CD}{sinA}

\Rightarrow \frac{AD . BD}{sin x sin (C - x)} = \frac{C D^{2}}{sinAsinB}

Given that

C D^{2} = AD . DB

Substitute this value of

C D^{2} = AD . DB

in RHS of the equation

\Rightarrow \frac{AD . BD}{sin x sin (C - x)} = \frac{AD . BD}{sinAsinB}

\Rightarrow \frac{1}{sin x sin (C - x)} = \frac{1}{sinAsinB}

Cross multiply both sides of the equation

\Rightarrow \frac{sinA}{sinB} = \frac{sinx}{\sin (C - x)}

We are also given that

sinAsinB = c^{2}

, substitute this value in LHS of the equation

\Rightarrow c^{2} = sin x sin (C - x)

Now divide and multiply RHS of the equation by 2 so as to use the identity

2 sinasinb

\Rightarrow c^{2} = \frac{1}{2} [2 sinxsin (C - x)]

We know that 2sinasinb=cos(a−b)−cos(a+b)

\Rightarrow c^{2} = \frac{1}{2}

[cos(x−C+x)−cos(x+C−x)]

\Rightarrow c^{2} = \frac{1}{2}

[cos(2x−C)−cosC]
Cross multiply the value of 2 to LHS of the equation
⇒

c^{2} =

cos(2x−C)−cosC
Use the formula cos2a=1−2sin2a in LHS i.e.

c^{2} =

1−cosC
⇒1−cosC=cos(2x−C)−cosC
⇒1=cos(2x−C)
We know

\cos 0 ° = 1

⇒cos0

°

=cos(2x−C)
Take inverse cosine on both sides of the equation
⇒cos−1(cos0

°

)=cos−1(cos(2x−C))
⇒0 = 2x−C
⇒2x = C
Divide both sides by 2

\Rightarrow x = \frac{C}{2}

Then ∠ACD=

\frac{C}{2}

and ∠BCD = C −

\frac{C}{2}

=

\frac{C}{2}

Since both angles are equal, so CD is the angle bisector of ∠C.
Hence proved.
So, the given statement is true.

Choose whether the given statement is true or false.

In a scalene triangle ABC, D is a point on the side AB such that CD2=AD.DB , if $sinAsinB = \frac{C}{2}$ then prove that CD is internal bisector of ∠C.

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