Question

In a scalene triangle ABC, D is a point on the side AB such that \[C{D^2} = AD.DB\], if \[\sin A\sin B = {\sin ^2}\dfrac{C}{2}\]then prove that CD is internal bisector of \[\angle C\]

Answer 1

Hint: Assume one of the bisected angles as a variable and write another angle in terms of assumed variable. We use sine rules to write equations for both the triangles. Multiply equations formed from two triangles and cancel all possible terms. Use the identity of \[2\sin a\sin b\] to solve the equation.
* Law of sine states that in a triangle ABC the ratio of side of a triangle to the sine of opposite angle is same for all the sides of the triangle, i.e. if side a has opposite angle A, side b has opposite angle B and side c has opposite angle C, then we can write \[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = k\] where k is some constant term.

Complete step-by-step solution:
We draw a scalene triangle ABC having point D on side AB.

Here \[\angle C\] is divided into two angles by the line CD i.e. \[\angle ACD\] and \[\angle BCD\].
Let \[\angle ACD = x\] then we can write \[\angle BCD = C - x\]
Now we apply sine rule in both triangles formed.
In \[\vartriangle ACD\],
\[ \Rightarrow \dfrac{{AD}}{{\sin x}} = \dfrac{{CD}}{{\sin A}}\]...............… (1)
In \[\vartriangle BCD\],
\[ \Rightarrow \dfrac{{BD}}{{\sin (C - x)}} = \dfrac{{CD}}{{\sin B}}\]................… (2)
Now we multiply both the equations (1) and (2)
\[ \Rightarrow \dfrac{{AD}}{{\sin x}} \times \dfrac{{BD}}{{\sin (C - x)}} = \dfrac{{CD}}{{\sin B}} \times \dfrac{{CD}}{{\sin A}}\]
Multiply numerator by numerator and denominator by denominator
\[ \Rightarrow \dfrac{{AD.BD}}{{\sin x\sin (C - x)}} = \dfrac{{C{D^2}}}{{\sin A\sin B}}\]
We are given that \[C{D^2} = AD.DB\]
Substitute this value of \[C{D^2} = AD.DB\]in RHS of the equation
\[ \Rightarrow \dfrac{{AD.BD}}{{\sin x\sin (C - x)}} = \dfrac{{AD.BD}}{{\sin A\sin B}}\]
Cancel same terms from both sides of the equation
\[ \Rightarrow \dfrac{1}{{\sin x\sin (C - x)}} = \dfrac{1}{{\sin A\sin B}}\]
Cross multiply both sides of the equation
\[ \Rightarrow \sin A\sin B = \sin x\sin (C - x)\]
We are also given that \[\sin A\sin B = {\sin ^2}\dfrac{C}{2}\], substitute this value in LHS of the equation
\[ \Rightarrow {\sin ^2}\dfrac{C}{2} = \sin x\sin (C - x)\]
Now divide and multiply RHS of the equation by 2 so as to use the identity \[2\sin a\sin b\]
\[ \Rightarrow {\sin ^2}\dfrac{C}{2} = \dfrac{1}{2}\left[ {2\sin x\sin (C - x)} \right]\]
We know that \[2\sin a\sin b = \cos (a - b) - \cos (a + b)\]
\[ \Rightarrow {\sin ^2}\dfrac{C}{2} = \dfrac{1}{2}\left[ {\cos (x - C + x) - \cos (x + C - x)} \right]\]
\[ \Rightarrow {\sin ^2}\dfrac{C}{2} = \dfrac{1}{2}\left[ {\cos (2x - C) - \cos C} \right]\]
Cross multiply the value of 2 to LHS of the equation
\[ \Rightarrow 2{\sin ^2}\dfrac{C}{2} = \cos (2x - C) - \cos C\]
Use the formula \[\cos 2a = 1 - 2{\sin ^2}a\]in LHS i.e. \[2{\sin ^2}\dfrac{C}{2} = 1 - \cos C\]
\[ \Rightarrow 1 - \cos C = \cos (2x - C) - \cos C\]
Cancel same terms from both sides of the equation
\[ \Rightarrow 1 = \cos (2x - C)\]
We know \[\cos {0^ \circ } = 1\]
\[ \Rightarrow \cos {0^ \circ } = \cos (2x - C)\]
Take inverse cosine on both sides of the equation
\[ \Rightarrow {\cos ^{ - 1}}\left( {\cos {0^ \circ }} \right) = {\cos ^{ - 1}}\left( {\cos (2x - C)} \right)\]
Cancel inverse function by the function
\[ \Rightarrow 0 = 2x - C\]
Shift C to RHS
\[ \Rightarrow 2x = C\]
Divide both sides by 2
\[ \Rightarrow x = \dfrac{C}{2}\]
Then \[\angle ACD = \dfrac{C}{2}\]and \[\angle BCD = C - \dfrac{C}{2} = \dfrac{C}{2}\]
Since both angles are equal, so CD is the angle bisector of \[\angle C\].
Hence proved

Note: Many students make mistake of considering the other values where cosine equals 1 but keep in mind if we take other values like \[2\pi \] then we will get the value of C as \[2\pi + 2x\] which cannot happen in a triangle as sum of all interior angles is equal to \[\pi = {180^ \circ }\].