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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.

N Sekhar , 6 Years ago
Grade 12th pass
anser 1 Answers
Aditya Gupta

Last Activity: 5 Years ago

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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