DE and QR are parallel lines interested by transversal n . Intersecting DE at A and QR at D .AP and DP are factors of angle EAD and angle RDA respectively fixed angle APD.
Answers
Given : CE and QR are parallel lines interested by transversal n . Intersecting CE at A and QR at D .AP and DP are bisectors of angle EAD and angle RDA respectively
To Find : ∠APD.
Solution:
CE and QR are parallel lines interested by transversal AD
Properties of angles formed by transversal line with two parallel lines :
• Corresponding angles are congruent.
• Alternate angles are congruent. ( Interiors & Exterior both )
• Interior angles are supplementary. ( adds up to 180°)
=> ∠EAD + ∠RDA = 180°
AP and DP are bisectors of ∠EAD and ∠RDA
=> ∠DAP = (1/2) ∠EAD
∠ADP = (1/2) ∠RDA
=> ∠DAP + ∠ADP = (1/2)(∠EAD + ∠RDA )
=> ∠DAP + ∠ADP = (1/2)(180° )
=> ∠DAP + ∠ADP = 90°
∠DAP + ∠ADP + ∠APD = 180° Sum of angles of triangle
=> 90° + ∠APD = 180°
=> ∠APD = 90°
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