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.

Question

Aditya Gupta · Answer

this is an EXTREMELY easy ques. lemme explain.
note that if angle CBD= p, then angle CDB= 90 – p= angle ADE.

since angle AED= 90 (given), so EAD= 180 – (90+90 – p)= p

let AE=a so that BD= 2a

now AC= AD+DC

now ADcosp=AE= a

or AD= asecp

also BDsinp= DC= 2asinp

so AC= AD+DC= asecp+2asinp

but AC= BC

and BC= BDcosp= 2acosp

so asecp+2asinp= 2acosp
or 1+2sinpcosp= 2cos^2p

or sin2p=cos2p

or tan2p=1=tan45

or 2p=45

or p= 22.5

since angle ABC= 45, so p= 45/2 is half of it. thence 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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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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