In the given figure, `angleABC=80^(@)` and `angleDEF=45^(@)`. The arms DE and EF of `angleDEF` cut BC at P and Q respectively, Prove that `PD||BA`.

In the given figure, `angleABC=80^(@)` and `angleDEF=45^(@)`. The arms

1.	In the given figure, `angleABC=80^(@)` and `angleDEF=45^(@)`. The arms DE and EF of `angleDEF` cut BC at P and Q respectively, Prove that `PD\|\|BA`.
Answer» Let `angleEPQ=x^(@)` We known that the sum of the angles of a triangle is `180^(@)`. `:. angleEPQ+anglePEQ+angleEQP=180^(@)` `implies angleEPQ+45^(@)+35^(@)=180^(@)` `implies angleEPQ=(180^(@)-80^(@))=100^(@)`. Now, EPD is a straight line. `:. angleEPQ+angleDPQ=180^(@) implies 100^(@)+angleDPQ=180^(@)` `implies angleDPQ=(180^(@)-100^(@))=80^(@)`. Thus, `angleABP=angleDPQ` [each equal to `80^(@)`] But, these are corresponding angles. Hence, `PD\|\|BA`.