Question

In, $\Delta ABC,\;AD$ is a perpendicular bisector of $BC$ in such a way $B - D - C$ happen prove that $AB = BC$.

Answer 1

Hint: Perpendicular bisector is the line segment which is perpendicular to the side of the triangle or cut the side into two equal parts i.e. it will pass through the mid point of the side of the triangle.

Complete step by step answer:
Let us solve this question by first constructing the triangle.
Given,
Line AD is the perpendicular bisector of BC.
So, $\angle ADC = \angle ADB = 90^\circ {\text{ }}.....{\text{(1)}}$
And line AD bisects line BC, as it is the perpendicular bisector
$BD = CD{\text{ }}.....{\text{(2)}}$

                   

From the above figure –
In $\Delta ABD\,and\,\Delta ACD,$
AD = AD…………………………..( the Common sides of the triangle)
$\angle ADB = \angle ADC$...................... [From the equation (1)]
BD = CD……………………………...[From the equation (2)]
Therefore, by SAS congruence rule - If any two sides and the included angle between the sides of the one triangle are equivalent to the corresponding two sides and the angle between the sides of the second triangle, then the two given triangles are said to be the congruent by SAS rule.

Hence, $\Delta ADC \cong \Delta ADB$

By Corresponding parts of the congruent triangles (CPCT) which states that if two or more triangles are taken then the corresponding angles and the sides of the given triangles are also congruent to each other.
$\therefore AB = AC$
Therefore, triangle ABC is an isosceles triangle in which AB = AC.
In, $\Delta ABC,\;AD$ is a perpendicular bisector of $BC$ in such way $B - D - C$ happen then $AB = BC$
Hence, the required solution is proved.

Note: Also, follow the different conditions of the congruence of the triangles to prove these types of solutions such as –
> SSS criteria (Side - Side - Side)
> SAS criteria (Side – Angle - Side)
> ASA criteria (Angle – Side – Angle)
> AAS criteria (Angle – Angle – Side)
> RHS criteria (Right angle – Hypotenuse – Side)