solution of it fast
Answers
Answer:
(i)
AB = AC [ ABC is isosceles ]
BD = CD [ DBC is isosceles ]
∠ABD = ∠ABC - ∠DBC
= ∠ACB - ∠DCB [ triangles ABC and DBC are isosceles ]
= ∠ACD
Therefore ΔABD ≅ ΔACD [ SAS ]
(ii)
∠ABP = ∠ACP [ ABC is isosceles ]
∠BAP = ∠CAP [ ABD and ACD are congruent ]
AB = AC [ ABC is isosceles ]
Therefore ΔABP ≅ ΔACP [ AAS ]
(iii)
∠BAP = ∠CAP [ ABD and ACD are congruent ]
=> AP bisects ∠A
∠BDP = ∠BAD + ∠DBA [ external angle of a triangle ]
= ∠CAD + ∠DCA [ ABD and ACD are congruent ]
= ∠CDP [ external angle of a triangle ]
=> DP (=AP) bisects ∠D
Therefore AP bisects both ∠A and ∠D
(iv)
BP = CP [ ABP and ACP are congruent ]
=> AP bisects BC
∠BPA = 180° - ∠CPA
= 180° - ∠BPA [ ABP and ACP are congruent ]
=> 2∠BPA = 180°
=> ∠BPA = 90°
=> AP is perpendicular to BC
Therefore AP is the perpendicular bisector of BC