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A B C is a triangle in which /B =2 ...

`A B C` is a triangle in which `/_B` =2 `/_CdotD` is a point on `B C` such that `A D` bisects `/_B A C` and `A B=C D` . Prove that `/_B A C=72^0dot`

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Let BP be the bisector of `angleABC`. Join PD.
Let `angleBAD=angleDAC=x`
Let `angleABP=anglePBC=y`
Also `angleB=2angleC`
`implies" "2y=2angleC`
`implies" "angleC=y`
Since `anglePBD=anglePCD=y`
`implies" "PB=PC` ...(I) (sides opposite to equal angles are equal)
Now, in `DeltaABP` and `DeltaDCP`
`:' {(AB=DC,("given")),(angleABP=angleDCP,("each equal to y")),(PB=PC,["from (1)"]):}`
`:. DeltaABP cong DeltaDCP` (SAS)
`:. angleBAP=angleCDP` (c.p.c.t)
`implies" "angleCDP=2x` and `AP=DP` (c.p.c.t)
`:. angle PDA=anglePAD=x`
For `DeltaABD, angleADC` is the exterior angle
`:. angleCDP+angleADP=x+2y" "("ext. "angleth.)`
`implies" "2x+x=x+2y" "implies" "x=y`
Now, in `DeltaABC`,
`angleA+angleB+angleC=180^(@)` (angle sum property)
`implies" "2x+2y+y=180^(@)`
`implies" "2x+2x+x=180^(@)" "( :' x=y)`
`implies" "5x=180^(@)" "implies" "x=36^(@)`
`implies" "angleBAC=2x=72^(@)` Hence Proved.
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