If ^@ABC^@ is a triangle and ^@D^@ is a point on side ^@AB^@ with ^@AD = BD = CD^@, find the value of ^@\angle ACB^@.


Answer:

^@90^\circ^@

Step by Step Explanation:
  1. It is given that ^@D^@ is a point on the side ^@AB^@ of a ^@\triangle ABC^@ such that ^@AD = BD = CD^@.

    AC DB

    We are required to find the value of ^@\angle ACB.^@
  2. 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}^@
  3. 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}^@
  4. Hence, the value of ^@\angle ACB^@ is ^@90^{ \circ }^@.

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