Global
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^@
- 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 }^@.
