In triangle ABC,AD perpendicular to BC,such that AD^2=BD*CD.Prove that triangle ABC is right angled at A.
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Proof:-
In right angled triangles ADB and ADC
BD^2+AD^2=AB^2………….(1)
DC^2++AD^2=AC^2………..(2)
On adding eq.(1) & (2)
BD^2+DC^2+2AD^2=AB^2+AC^2
Put AD^2=BD.DC (given)
BD^2+DC^2+2BD.DC=AB^2+AC^2
(BD+DC)^2=AB^+AC^2 , put BD+DC=BC
BC^2=AB^2+AC^2 , therefore in triangle ABC
BC is hypotenues and angle BAC=90° , Proved.
In right angled triangles ADB and ADC
BD^2+AD^2=AB^2………….(1)
DC^2++AD^2=AC^2………..(2)
On adding eq.(1) & (2)
BD^2+DC^2+2AD^2=AB^2+AC^2
Put AD^2=BD.DC (given)
BD^2+DC^2+2BD.DC=AB^2+AC^2
(BD+DC)^2=AB^+AC^2 , put BD+DC=BC
BC^2=AB^2+AC^2 , therefore in triangle ABC
BC is hypotenues and angle BAC=90° , Proved.
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