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In the given figure, angleABC=80^(@) and...

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`.

Text Solution

Verified by Experts

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`.
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