Question

A line drawn from vertex A of an equilateral triangle ABC meets BC at D and circumcircle at P
show that : PA = PB

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Answer 1

Step-by-step explanation:

Given A line drawn from vertex A of an equilateral triangle ABC meets BC at D and circumcircle at P show that : PA = PB + PC

• According to the question ABC is an equilateral triangle where BC meets at a point D and circumcircle at a point P. We need to find              PA = PB + PC
• Now from the triangle AB = BC = CA
• Since angles are in same segment,
• So angle ABC = angle ACP = 60 degree
• Also angle ACB = angle ABP = 60 degree
• Now consider the triangle ACP,
• We have by cosine rule c^2 = a^2 + b^2 – 2ab cos c
• Or we can write it as cos c = a^2 + b^2 – c^2 / 2ab
• Or cos 60 = AP^2 + CP^2 – AC^2 / 2 (AP)(CP)
•        1/2  = Ap^2 + CP^2 – AC^2 / 2(AP)(CP)
• Or AP^2 – AC^2 = (AP)(CP) – CP^2 -----------1
• Now in triangle APB we have
•      Cos 60 = AP^2 + BP^2 – AB^2 /  2 (AP)(BP)
•        1/2  = AP^2 + BP^2 – AB^2 /  2 (AP)(BP)
• Or AP^2 – AB^2 = (AP)(BP) – BP^2
• Or AP^2 – AC^2 = (AP)(BP) – BP^2 ------------2
• From equation 1 and 2 we get
• (AP)(PC) – PC^2 (AP)(PB) – PB^2
• Or PC^2 - PB^2 = (AP)(PC) – (AP)(PB)  
• Or (PC + PB)(PC – PB) = AP(PC – PB) (since a^2 - b^2 = (a + b)(a – b)
• Or AP(PC – PB) = (PC + PB)(PC – PB)
• Or AP = (PC + PB)(PC – PB) / (PC – PB)
• Or AP = PC + PB
• So PA = PB + PC

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https://brainly.in/question/14874177