Скачать или смотреть theorem 12.1.2 class 9

Theorem 12.1.2
(converse of Theorem 12.1.1)
Any point equidistant from the end points of line segment is on the right bisector of it.
Given
¯𝑨𝑩 is a line segment. Point P is such that ¯𝑷𝑨 ≅ ¯( 𝑷𝑩).
To Prove
The point P is on the right bisector of ¯𝑨𝑩
Construction: joint P to C the mid-point ¯𝑨𝑩 .
In ΔACP ⟷ ΔBCP
¯( 𝑃𝐴) ≅ ¯( 𝑃𝐵)
¯𝑃𝐶 ≅ ¯( 𝑃𝐶)
¯𝐴𝐶 ≅ ¯( 𝐵𝐶)
ΔACP ≅ ΔBCP
∠𝐴𝐶𝑃 "≅" ∠BCP ….i
But
m∠𝐴𝐶𝑃"+m"∠BCP=180o ….ii
𝑚∠𝐴𝐶𝑃=𝑚∠BCP=90o
(𝑃𝐶) ⃡ ⊥ (𝐴𝐵) ⃡ and ¯𝐶𝐴 ≅ ¯( 𝐶𝐵)
(𝑃𝐶) ⃡ is a right bisector of (𝐴𝐵) ⃡
i.e. the point P is on the right bisector of (𝐴𝐵) ⃡