Question

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.​

Answer 1

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.

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Answer 2

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= BD•cosx .....(1)

and AE = AD•cosx

But , AD = AC – CD = k – BD•sinx

So, that

AE= (k – BD•sinx)•cosx

or

2AE= BD= 2(k – BD•sinx)•cosx

.........(2)

Put k from (1) in (2),

BD= 2•cosx(BD•cosx – BD•sinx)

or

1 = 2•cosx(cosx – sinx)

or

2cos^2x – 1= 2sinxcosx

or

cos•2x = sin•2x

or

tan2x = 1 = tan•45

or

x = 22.5, which is obviously half of angle ABC (since angle ABC= 45 deg).

Hence, BD bisects angle ABC.

Thus, proved.