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Step-by-step explanation:

to \: prove :- \\ angle \: apc = 90 \: degrees

so \: here \: ap \: is \: bisector \: of \: angle \: bac \: and \: cp \: is \: bisector \: of \: angle \: dca \\ thus \: by \: angle \: bisector \: theorem \\ we \: get \\ angle \: bap = angle \: pac = 1/2 \times angle \: bac \\ ie \: angle \: pac = 1/2 \times \: angle \: bac \\ so \: angle \: bac = 2.angle \: pac \: \: \: \: \: \: \: \: \: \: \: \: \: \: (1) \\ \\ similarly \: \: angle \: dcp = angle \: pca = 1/2 \times angle \: dca \\ ie \: angle \: pca = 1/2 \times angle \: dca \\ so \: angle \: dca = 2.angle \: pca \: \: \: \: \: \: \: \: \: \: \: \: (2)

so \: as \: ab || cd \\ by \: interior \: angle \: theorem \\ we \: get \\ angle \: bac + angle \: dca = 180 \\ 2.angle \: pac + 2.angle \: pca = 180 \: \: \: \: \: \: \: \: \: \: \: \: \: \: from \: (1) \: and \: (2)

ie \: 2(angle \: pac + angle \: pca) = 180 \\ so \: thus \: then \\ angle \: pac + angle \: pca = 90 .\: degrees \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (3)

now \:by \: applying \: angle \: sum \: property \: on \: triangle \: apc \\ we \: get \\ angle \: apc + angle \: pac + angle \: pca = 180 \\ ie \: angle \: apc + 90 = 180 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: from \: (3) \\ ie \: angle \: apc = 90 \: degrees \\ \\ hence \: proved


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