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Definition | Figure |
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Altitude is the perpendicular line drawn from a vertex to the opposite side. AD ⊥ BC, BE ⊥ AC, CF ⊥ AB Orthocentre is the intersection point of the 3 altitudes. Property: Angles formed by a side of the triangle at the orthocentre and the 3rd vertex are supplementary ∠BOC + ∠BAC = 180° ∠COA + ∠CBA = 180° ∠AOB + ∠ACB = 180° |
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Perpendicular Bisector is a perpendicular line that bisects a side of a triangle. Their intersection point is called the Circumcentre. O is also the centre of the Circumcircle of △ABC, which passes through A, B and C. OA, OB and OC are the radii of the circumcircle. Properties of perpendicular bisectors are covered in Circles Lesson. |
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Angle Bisector is a line that bisects an interior angle.
Their intersection point is called the Incentre.
O is also the centre of the Incircle of △ABC, which touches the sides AB, BC and CA at P, Q and R respectively. Property 1: As AB, BC and CA are the tangents, OP ⊥ AB, OQ ⊥ BC, OR ⊥ CA. Property 2: An angle bisector is equidistant from the lines that bound the bisected angle as shown in Figure 2. In this figure, the angle bisector is AD and XP = YP. This holds good for the other angle bisectors as well. Property 3: ∠BOC = 90° + 21 ∠BAC Property 4: OPAO=BCAB+AC |
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Median is a line drawn from a vertex to the midpoint of its opposite side. Their intersection point is called the Centroid. Property 1: A median divides the area of a triangles into 2 halves, ie, Ar(△ABD) = Ar(△ACD) = Ar(△CAF) = Ar(△CBF) = Ar(△BAE) = Ar(△BCE) Property 2: The centroid divides each median in the ratio of 2:1, ie, AK : KD = BK : KE = CK : KF = 2:1 Property 3: Ar(△AKE) = Ar(△AKF) = Ar(△BKF) = Ar(△BKD) = Ar(△CKD) = Ar(△CKE) = 61× Ar(△ABC) Property 4: Ar(△AFE) = Ar(△CED) = Ar(△BFD) = Ar(△DEF) = 41× Ar(△ABC) |
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Let the area of each smaller triangle be 1 unit. ∴ Area of △ ABC = 6 units Area(CDOE) = Area(△ COD) + Area( △ COE) = 2 units Ratio of areas of CDOE to △ ABC = 2 : 6 = 1 : 3 |
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∠OCD =∠OCA = 35° (CO is the ∠ bisector) ∠ADC =180° − ∠ADB =80° (Linear Pair) In △ ADC, ∠DAC + 80° + 70° =180° ⇒ ∠DAC =30° ∠BOC =90° + 21 ∠BAC (Angle bisector Property) =90°+ ∠DAC =120° |
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Lets start with the area of the innermost triangles and use a variable x. Area ( △ ADF): Area (△ AFE) = 1 : 2 ∴ Let the actual areas of △ ADF and △ AFE be x and 2x respectively. Area of △ ADE =x+2x=3x Area (△ ADE) : Area (△ CDE) = 2 : 3 Area(△CDE)3x=32 ⇒ Area(△CDE)=4.5x |
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