`DeltaABC` is a right triangle right angled at A such that `AB = AC` and bisector of `/_C` intersects the side AB at D. Prove that `AC + AD = BC`.

Explanation

Given in right angled ` Delta ABC ` AB=AC and CD is the bisector of `angleC`
Construction Draw `DE_|_BC.`
To prove AC+AD=BC
Proof In right angled `Delta ABC ,`AB =AC and BC is a hpotenuse [given]
`therefore angle A 90^(@)`
In `Delta DAC and Delta DEC" " angle A =angle 3=90^(@)`

` angle 1 =angle 2 ` [given ,CD is the bisector of `angle C` ]
DC=DC [common sides]
`therfore Delta DAC cong Delta DEC ` [by AAS cogruence rule ]
`rArr DA=DE ` [ by CPCT ] ....(i)
and AC =EC ...(ii)
In `Delta ABC AB=AC `
`angle C = angel B`
[angle opposite to equal sides are equal ] ...(iii)
Again in `Delta ABC angle A + angel B + angel C =180^(@)`
[by angle sum property of a triangel ]
`rArr 90^(@)+angle B+ angel B =180^(@)`
`rArr 2angel B=180^(@)-90^(@)`
`rArr 2angle B=90^(@)`
`rArr angle B=45^(@)`
in `Delta BED angle 5 =180^(@)-(angle B + angel 4 )`
[by angle sum property of a triangle ]
`=180^(@)-(45^(@)+90^(@)`
`=180^(@)-135^(@)=45^(@)`
` therefore angle B =agnel 5 `
`rArr DE=BE [because ` sides opposite to equal angles are equal ]
In From Eqs. (i) and (iv)
DA=DE=BE
BC=CE +EB
=CA +DA [from Eqs (ii) and (V)]
`therefore AD+AC=BC`