Question

∆ABC is the right angled triangle at B and the perpendicular drawn to the opposite side bisectit at D AD=DC=5cm find BD ​

Answer 1

This is your answer.

Answer 2

in ∆ABD & ∆BDC,
AD=DC
<BDA = <BDC=90°
BD=BD (common)
Therefore, ∆ABD ≈∆CBD ("≈" is for congruent)
AB=BC (by c.p.c.t)
let AB=BC=x
now applying Pythagoras theorem in ∆ABC,
x²+x²=(10)²
2x²=100
x²=50
x= 5√2=AB
Now applying Pythagoras theorem in ∆ABD,
(AB)²=(BD)²+(AD)²
=>(BD)²=(AB)²-(AD)²
=>(BD)²=(5√2)²-(5)²
=> (BD)²=50-25
=> (BD)²=25
=> BD=5cm