If ^@ABC^@ is a triangle and ^@D^@ is a point on side ^@AB^@ wit | Math Question and Answer

Answer:

$90^\circ$

Step by Step Explanation:

It is given that $D$ is a point on the side $AB$ of a $\triangle ABC$ such that $AD = BD = CD$ .
We are required to find the value of $\angle ACB.$
We are given,
$\begin{align} &AD = CD \\ \implies & \angle DAC = \angle DCA \text{ (Angles opposite to equal sides of a triangle) } && \ldots(1) \space\space\space\space\space\space \end{align}$
Also,
$\begin{align} &BD = CD \\ \implies & \angle DBC = \angle DCB \text{ (Angles opposite to equal sides of a triangle) } && \ldots(2) \space\space\space\space\space\space \end{align}$
In $\triangle ABC$ ,
$\begin{align} & \angle BAC + \angle ACB + \angle CBA = 180^{ \circ } && \text{ [Angle sum property of a Triangle]} \\ \implies & \angle DAC + \angle ACB + \angle DBC = 180^{ \circ } \\ \implies & \angle DCA + \angle ACB + \angle DCB = 180^{ \circ } && \text{ [By eq (1) and (2)]}\\ \implies & \angle ACB + \angle ACB = 180^{ \circ } \\ \implies & 2\angle ACB = 180^{ \circ } \\ \implies & \angle ACB = 90^{ \circ } \\ \end{align}$
Hence, the value of $\angle ACB$ is $90^{ \circ }$ .

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