P is a point on the side B C of a tr...

`P` is a point on the side `B C` of a triangle `A B C` such that `A B=A Pdot` Through `A\ a n d\ C ,\ ` lines are drawn parallel to `B C\ a n d\ P A ,` respectively, so as to intersect at `D` as shown in Figure. Show that `A B C D` is a cyclic quadrilateral. `

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In`/_\ABP`
`AB =AP`
`/_APB=/_ABP`(Angles opposing to equal sizes are equal)
`/_BPA=/_APC=180^@`(Linear pair)
`/_APC=180^@ - /_BPA` ........`(1)`
`/_PAD+ /_APC =180^@`(co interrior angles)
`/_PAD=180^@ - /_APC=180^@ - 180^@ + /_BPA`
`In/_\ABP`
`AB =AP`
`/_APB=/_ABP`(Angles opposing to equal sizes are equal)
`/_BPA=/_APC=180^@`(Linear pair)
`/_APC=180^@ - /_BPA` ........`(1)`
`/_PAD+ /_APC =180^@`(co interrior angles)
`/_PAD=180^@ - /_APC=180^@ - 180^@ + /_BPA`
`/_PAD=/_BPA .......(ii)`
`/_PAD + /_ACD =180^@`(co interior `/_s`)
`/_ADC=180^@ - /_PAD = 180^@ - /_BPA`.......(from (`ii`))
`/_ADC + /_BPA= 180^@`
`In/_\ABP`
`AB =AP`
`/_APB=/_ABP`(Angles opposing to equal sizes are equal)
`/_BPA=/_APC=180^@`(Linear pair)
`/_APC=180^@ - /_BPA` ........`(1)`
`/_PAD+ /_APC =180^@`(co interrior angles)
`/_PAD=180^@ - /_APC=180^@ - 180^@ + /_BPA`
`In/_\ABP`
`AB =AP`
`/_APB=/_ABP`(Angles opposing to equal sizes are equal)
`/_BPA=/_APC=180^@`(Linear pair)
`/_APC=180^@ - /_BPA` ........`(1)`
`/_PAD+ /_APC =180^@`(co interrior angles)
`/_PAD=180^@ - /_APC=180^@ - 180^@ + /_BPA`
`/_PAD=/_BPA .......(ii)`
`/_PAD + /_ACD =180^@`(co interior `/_s`)
`/_ADC=180^@ - /_PAD = 180^@ - /_BPA`.......(from (`ii`))
`/_ADC + /_BPA= 180^@`
`/_ADC + /_ABP= 180^@`( `/_BPA= /_ABP`)
Since sum of opposite angles of the quadilateral is `180^@`
Hence the quadilateral `ABCD` is a cyclic quadilateral.

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