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Given that ABC is an isosceles right triangle with AC=BC and angle ACB=90°.D is a point on AC and E is point on the extension of BD such that AE is perpendicular to BE. If AE=1/2BD,prove that BD bisects angle ABC.
let angle CBD= x. also, AC= BC= know, by angle chasing you can prove that angle DAE= angle CBD= xnow, BC= k= BDcosx.....(1)and AE= ADcosx. but AD= AC – CD= k – BDsinxso that AE= (k – BDsinx)cosxor 2AE= BD= 2(k – BDsinx)cosx.........(2)put k from (1) in (2)BD= 2cosx(BDcosx – BDsinx)or 1= 2cosx(cosx – sinx)or 2cos^2x – 1= 2sinxcosxor cos2x= sin2xor tan2x= 1= tan45or x= 22.5, which is obviously half of angle ABC (since angle ABC= 45 deg)hence, BD bisects angle ABC.kindly approve :)
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