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Constructions (NIB)

Constructions (NIB) - angles Therefore AP ←-→ and BP...

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Constructions (NIB) Tangent Construction (NIB): To construct a tangent to a given circle through a given point outside the circle. Given: Circle having center O and point P outside the circle. To be constructed: A tangent to the circle containing point P . Construction: Step 1: Construct segment OP . Step 2: Construct the midpoint of segment OP and label this point M. Step 3: Construct the circle C( M, MP ) . Label the two points where the two circles intersect A and B. Step 4: Construct lines AP ←-→ and BP ←-→ . Lines AP ←-→ and BP ←-→ are tangent to circle with center O and they contain point P. Proof: C( M, MP ) contains point O because OM = MP, since M is the midpoint of OP . Thus, OP is a diameter of circle C( M, MP ). Therefore, each of the two angles, OAP and OBP, is inscribed in a semi-circle. Therefore, by Corollary 4.5.12, OAP and OBP are right

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Unformatted text preview: angles. Therefore, AP ←-→ and BP ←-→ are tangent to the circle with center O by Theorem (NIB) 4.4 . Q E D A B M O P O P Construction (NIB): Given two points P and Q, construct the set of all centers of circles which contain both P and Q . Solution: Construct the perpendicular bisector of the segment . Given: Construct: Construction (NIB): Given a line m and a point P on line m, construct the set of the centers of all of the circles to which line m is a tangent line with point of tangency at point P . Solution: Construct the line through P which is perpendicular to line m . Given: Construct: Q P m P m P Q P O 1 O 2 O 3 Construction (NIB): Given a line m which contains point P and given another point Q, with Q not on line m, construct a circle passing through Q and tangent to line m at point of tangency P. Given: Construct: Q m P Z Q m P...
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Constructions (NIB) - angles Therefore AP ←-→ and BP...

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