Question

in the figure angle ABC=66 and angle angle DAC= 38 degree ce is perpendicular to ab and ad is perpendicular to bc prove that cpis greater than ap​

Answer 1

Proved that CP >   AP    if ∠ABC = 66° & ∠DAC = 38°  , CE⊥ AB  & AD ⊥BC  & AD & CE intersect at P

Step-by-step explanation:

P is the point of intersection of  AD & CE

∠DAC = 38°

=> ∠PAC = 38°  as P lies on DA

Now in Δ ADC

∠DAC + ∠ADC + ∠DCA  = 180°

=> 38° + 90°  + ∠DCA  = 180°

=> ∠DCA  = 52°

Now in Δ BCE

∠EBC = ∠ABC = 66°

∠BEC =  90°

∠EBC + ∠BEC + ∠BCE = 180°

=> 66° + 90° + ∠BCE = 180°

=> ∠BCE = 24°

∠DCP = ∠BCE = 24°    

∠PCA = ∠DCA - ∠DCP

=> ∠PCA =  52° - 24°

=> ∠PCA =  28°

Now in Δ APC

∠PAC = 38°

∠PCA =  28°

=>∠PAC > ∠PCA

=> CP >   AP  ( as Side opposite to greater angle is Greater)

QED

Proved

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Answer 2

Answer:

nice post thanks for the first answer