Question

In the given figure ∠ABC = 66°, ∠DAC = 38°. CE is perpendicular to AB and AD is perpendicular to BC. Prove that : CP > AP

Answer 1

in Quadrilateral EBDP formed

∠PEB + ∠EBD + ∠BDP + ∠DPE  =  360°

∠PEB  = 90° as CE⊥ AB  & P is point on CE

∠EBD = ∠ABC = 66°    (as E & D are points of AB & BC)

∠BDP  90° as AD⊥ BC  & P is point on AD

=> 90° + 66° + 90° + ∠DPE  =  360°

=>  ∠DPE  = 114°

∠APC =  ∠DPE    ( opposite angles)

=> ∠APC = 114°

in Δ APC

∠PAC  + ∠PCA + ∠APC = 180°

∠PAC = ∠DAC = 38°

=> 38°  + ∠PCA + 114° = 180°

=> ∠PCA = 28°

∠PAC  >  ∠PCA   ( as 38°  > 28° )

=> CP > AP    as side opposite to greater angle is greater

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