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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.`
8 months 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)= plet AE=a so that BD= 2anow AC= AD+DCnow ADcosp=AE= aor AD= asecpalso BDsinp= DC= 2asinpso AC= AD+DC= asecp+2asinpbut AC= BCand BC= BDcosp= 2acospso asecp+2asinp= 2acospor 1+2sinpcosp= 2cos^2por sin2p=cos2por tan2p=1=tan45or 2p=45or p= 22.5since angle ABC= 45, so p= 45/2 is half of it. thence BD bisects angle ABC.kindly approve :)
```
7 months ago
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