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Properties of Isosceles and Equilateral Triangles

In an isoscel...

Question

In an isosceles $△ A B C$ , if $A B = A C$ and $D$ is a point on $B C$ , then prove that
${A B}^{2} - {A D}^{2} = B D . C D$

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Solution

Given: $A B = A C$
In triangle $A B C$
$cos c = \frac{{A C}^{2} + {B C}^{2} - {A B}^{2}}{2 A C . B C}$
$= \frac{{B C}^{2}}{2 A C . B C} = \frac{B C}{2 A C} \to i$
In $△ A D C$
$cos C = \frac{{A C}^{2} + {D C}^{2} - {A D}^{2}}{2 A C . D C} \to i i$
$I$ and $i i$
$\frac{B C}{2 A C} = \frac{{A C}^{2} + {D C}^{2} - {A D}^{2}}{2 A C . B C}$
$\Rightarrow B C . D C = {A C}^{2} + {D C}^{2} - {A D}^{2}$
$(B D + D C) D C = {A B}^{2} + {D C}^{2} - {A D}^{2} = B D . D C = {A B}^{2} - {A D}^{2}$

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