Q:2. In a isosceles triangle ABC, AB = BC and AD is perpendicular on BC .So prove that BD=CD
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Answer:
∵ACB is isos. Δ,
AO⊥BCαBO=CO :
T.P.:AD
2
=AC
2
+BD×CD
Proof :
AD
2
=AO
2
+OD
2
AB
2
=AO
2
+BO
2
}
Pythogearus....(i)
theorem....(ii)
Subtract (ii) from (i)
AD
2
−AB
2
=OD
2
−BO
2
AD
2
−AB
2
=(OD+BO)(OD−BO)[∵a
2
−b
2
=(a+b)(a−b)]
AD
2
−AB
2
=(BD)(OD−OC)[∵BO=CO]
AD
2
−AB
2
=(BD)(CD)[∵OD−OC=CD]
AD
2
=AB
2
+BD×CD
AD
2
=AC
2
+BD×CD[∵AB=AC&AB
2
=AC
2
]
Hence , proved
Step-by-step explanation:
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