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

In △ABD,

By angle sum property of triangle,

⇒ ∠ABD + ∠ADB + ∠BAD = 180°

⇒ 66° + 90° + ∠BAD = 180°

⇒ ∠BAD + 156° = 180°

⇒ ∠BAD = 180° - 156°

⇒ ∠BAD = 24°

In △AEP,

By angle sum property of triangle,

⇒ ∠AEP + ∠APE + ∠BAD = 180°

⇒ 90° + ∠APE + 24° = 180°

⇒ ∠APE + 114° = 180°

⇒ ∠APE = 180° - 114°

⇒ ∠APE = 66°

From figure,

∠APE + ∠APC = 180° (Linear pair)

⇒ 66° + ∠APC = 180°

⇒ ∠APC = 180° - 66°

⇒ ∠APC = 114°

In △APC,

By angle sum property of triangle,

⇒ ∠APC + ∠ACP + ∠PAC = 180°

⇒ 114° + ∠ACP + 38° = 180°

⇒ ∠ACP + 152° = 180°

⇒ ∠ACP = 180° - 152°

⇒ ∠ACP = 28°

We know that,

The shortest side of a triangle has the shortest angle opposite to it.

⇒ AP < CP

⇒ CP > AP.

Hence, proved that CP > AP.