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Grade: 9
        
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.
5 months ago

Answers : (1)

Aditya Gupta
1717 Points
							
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 :)
3 months ago
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