Answer :


Given that ΔABC and ΔDBC are isosceles triangles.


Here, AB=AC and DB=DC


Also, ABC=ACB


And DBC=DCB


Now, in ΔABD and ΔACD,


AD=AD (common)


AB=AC (given)


BD=DC (given)


ΔABD ΔACD (by SSS rule)


BAD=CAD (by cpct)


Again, in ΔABO and ΔACO,


AO=AO (common)


BAO=CAO (as BAD=CAD)


AB=AC (given)


ΔABD ΔACD (by SAS rule)


So, BO=OC (by cpct)


And AOB=AOC(by cpct)


Also, AOB+AOC=180° (straight angle)


AOB+AOB=180°


2AOB=180°


AOB=90°


So, AOC=AOB=90°


Hence, AD bisects BC perpendicularly.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

In trapezium ABCDWest Bengal - Mathematics

Let’s see the isoWest Bengal - Mathematics

Let’s see the isoWest Bengal - Mathematics

let’s see the isoWest Bengal - Mathematics

let’s see the isoWest Bengal - Mathematics

E and F are twp pWest Bengal - Mathematics

Two line segmentsWest Bengal - Mathematics

Two isosceles triWest Bengal - Mathematics

Let’s see the isoWest Bengal - Mathematics

If we produce theWest Bengal - Mathematics