Answer :
We have ⊿ ABC is an equilateral triangle and AD⊥BC
In ⊿ ADB⊿ ADC
∠ADB=∠ADC=90°
AB=AC (Given)
AD=AD (Common)
⊿ ADB≅⊿ ADC (By RHS condition)
∴ BD=CD=BC/2 ……. (i)
In ⊿ ABD
BC2=AD2+BD2
BC2=AD2+BD2 [Given AB=BC]
(2BD)2= AD2+BD2 [From (i)]
4BD2-BD2=AD2
AD2=3BD2
Rate this question :
How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
BPT ke Master Kaise Bane?46 mins
Similarity - Apply it like a Champ45 mins




Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
view all courses
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation

