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ABC is a triangle with AB = AC. D is the Mid-point of BC .
prove that angle ADB is a right angle

Answer 1

$\textbf{ \underline{ \underline{Solution : \: \: } \: }} \\ \\ </p><p> \textsf{ \underline{ Given : }} \\ \\ \star \: \text{ \small{ABC \: is \: a \: triangle}} \\ \star \: \text{ \small{AB = AC }} \\ \star \: \text{ \small{D is the mid point of BC }} \\ \\ </p><p></p><p> \textsf{ \underline{ To Prove : }} \\ \\ \star \: \text{ \small{ \angle ADB \: is \: a \: right \: angle }} \\ \\ </p><p>\boxed{ \textsf{(Refer the attachment for figure) }}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \text {\underline{ \: BC is a straight line \: }} \\ \implies\text{\small{ \angle ADB + \angle ADC = 180 \degree }} \: \: \: \bigg( \text{Angles on the line BC}\bigg) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \text{AB = AC} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg( \text{Given}\bigg) \\ \implies \: \text{\small{ \angle ADB = \angle ADC}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg( \textbf{Theorem : } \text{Opposite angle of two equal lines are also equal}\bigg) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \textbf{\small{ \angle ADB + \angle ADC = 180 \degree }} \\ \implies \: \text{ \small{ \angle ADB + \angle ADB = 180 \degree }} \\ \implies \: \text{ \small{2 \angle ADB = 180 \degree }} \\ \implies \: \text{ \small{ \angle ADB}} = \frac{180\degree }{2} \\ \implies \: \text{ \small{ \angle ADB}} = 90\degree \\ \\ \boxed{ \boxed{ \textsf{Hence, \angle ADB is a Right angle }}}$