In the given figure, `DeltaABC` is right angled at B such that `angleBCA=2angleBAC.` Show that hypotenuse AC = 2BC.

Explanation

GIVEN A `DeltaABC` in which `angleB=90^(@) and angleBCA=2angleBAC.`
TO PROVE AC = 2BC.
CONSTRUCTION Produce CB to D such that BD = BC.
Join AD.
PROOF Let `angleBAC=x^(@)." Then, "angleBCA=2x^(@).`
In `DeltaABC and DeltaABD,` we have
`BC=BD" (by construction)"`
`AB=AB" (common)"`
`angleABC=angleABD=90^(@)`
`therefore" "DeltaABC~=DeltaABD" (SAS-criteria)"`
`therefore" "angleCAB=angleDAB=x^(@)" ...(i) (c.p.c.t.)."`
and `AC=AD" ...(ii) (c.p.c.t.)."`
In `DeltaCAD`, we have
`angleCAD=angleCAB+angleDAB=(x^(@)+x^(@))=2x^(@)`.
`angleACD=angleACB=2x^(@).`
But, we know that the sides opposite to equal angles are equal.
`therefore" "angleACD=angleCAD rArr AD=CD." ...(iii)"`
From (ii) and (iii), we get
`AC=CD rArr AC=2BC" "[because CD=BC+BD=2BC].`
Hence, `AC=2BC.`