In the figure, ABD, BCD are two triangles. BD is the bisector of `angleABC and angleADC. bar(AB)" || "bar(CD) and bar(AD)" || " bar(BC)`. If `angleBCD=50^(@)`, then find the angle of `angleBAD`.

Explanation

`angleBCD=50^(@)and bar(BC)" || "bar(AD)`.
`angleCDA+angleBCD=180^(@)" (co-interior and angles)"`
`rArr angleCDA+50^(@)=180^(@)`
`angleCDA=180^(@)-50^(@)`
`angleCDA=130^(@)`
Similarly, `angleABC=130^(@)" "(because bar(AB)" || "bar(CD))`
Since BD is the bisector of `angleADC and angleABC`,
`angleABC=angleABD+angleCBD" "(because angleABD=angleCBD)`
`2angleABD=130^(@)rArr angleABD=angleCBD=(130)/(2)=65^(@)`
`angleADB+angleBDC=130^(@)and angleADB=angleBDC=(130)/(2)=65^(@)`
`lnDeltaABD,angleABD+angleADB+angleBAD=180^(@)`
`130^(@)+angleBAD=180^(@)`
`angleBAD=180^(@)-130^(@)=50^(@)`