Question

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

Answer 1

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 )
therefore,
 ΔBEC is isosceles        ( because by the property of isosceles triangle base angles are equal  ) 

Answer 2

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.