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Given :In Δ ABC: BD is perpendicular to AC, CE is perpendicular to AB and BD=CE.Prove:ΔABC is isosceles. refer to the pic

TO PROVE ABC is isocels triangle ?


given , 
    BD = CE 
  angle BEC = angle BDC = 90^0
To prove -  ΔBEC is congruent to ΔBDC
proof ,
    BD = CE ( Given )
     angle BEC = angle BDC  ( Given )
     BC = BC ( common )
 by SAS  ΔBEC is congruent to ΔBDC ,
therefore , angle EBC = DCB ( C.P.C.T )
 ΔBEC is isosceles        ( because by the property of isosceles triangle base angles are equal  ) 

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Given: EC perpendicular AB &BD Perpendicular AC BD=CE TO PROVE:to prove triangle ABC is an isosceles triangle proof:in triangle EBC and triangle DCB, - BD=CE (Given) -/_CEB=/_BDC=90°(EC Perpendicular AB & BD perpendicular AC) BC=CB(common) Hence, triangle EBC=~ triangle DCB( BY RHS) THEREFORE, /_EBC =/_DCB(BY CPCT) I.e., /_ABC=/_ACB Hence, triangle ABC is an isosceles triangle.
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