In triangle ABC, angle ABC = 66, CE is perpendicular to AB and AD is perpendicular to BC and CE and AD intersect at point P. Angle DAC = 38, prove that CP > AP.
Answers
Proved that CP > AP in triangle ABC ∠ABC = 66° , CE ⊥ BP & AD ⊥BC intersect at P and ∠DAC = 38°
Step-by-step explanation:
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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Answer:
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