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.

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Aditya Gupta · Answer

let angle CBD= x. also, AC= BC= k now, by angle chasing you can prove that angle DAE= angle CBD= x now, BC= k= BDcosx.....(1) and AE= ADcosx. but AD= AC – CD= k – BDsinx so that AE= (k – BDsinx)cosx or 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= 2sinxcosx or cos2x= sin2x or tan2x= 1= tan45 or 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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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.

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.

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