Question

Please solve q 80 ?

Answer 1

option (4) is acceptable.

Explanation:

Given that, `AB=AC=BD` and `AC_|_BD`.

`rarrAB=AC`

`rarr/_B=/_C`

`rarr90-a+90-d=d`

`rarra=180-2d`.....[1]

Also, `rarrAB=BD`

`rarr/_A=/_D`

`rarra+b=90-b`

`rarra=90-2b`....[2]

From [1] and [2], we have,

`rarr180-2d=90-2b`

`rarrd-b=45`....[3]

Now, `/_C+/_D`

`=/_BCA+/_BDA=90-b+d=90+45=135`

Answer 2

`angleC+angleD=135^@`

Explanation:

Let `angleABD=x`,
given `AB=BD, => angleBDA=(180-x)/2`
let `angleCBD=y`,
given `AB=AC, => angleACB=angleABC=x+y`,
In `DeltaCBE, => color(red)(x+2y=90^@)`
`angleC+angleD=angleACB+angleBDA=x+y+(180-x)/2`
`=x+y+90-x/2`
`=x/2+y+90`
`=(x+2y)/2+90`
`=90/2+90=45+90=135^@`