Question

In the given figure, angleABC=66°, angleDAC=38°.CE is perpendicular to AB and BC. Prove that : CP > AP

Answer 1

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