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CAT 2025 Lesson : Lines & Triangles - Triangle & its Segments

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3.3 Altitudes, Medians, Angle & Perpendicular Bisectors

Definition Figure
Altitude is the perpendicular line drawn from a vertex to the opposite side. AD \perp BC, BE \perp AC, CF \perp AB
Orthocentre is the intersection point of the
33 altitudes.

Property: Angles formed by a side of the triangle at the orthocentre and the
3rd3^{\mathrm{rd}} vertex are supplementary
\angleBOC + \angleBAC = 180°180\degree
\angleCOA + \angleCBA = 180°180\degree
\angleAOB + \angleACB = 180°180\degree
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 \triangleABC, 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.

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 \triangleABC, 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
\perp AB, OQ \perp BC, OR \perp 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:
\angleBOC = 90°90\degree + 12\dfrac{1}{2} \angleBAC

Property 4:
AOOP=AB+ACBC\dfrac{\mathrm{AO}}{\mathrm{OP}} = \dfrac{\mathrm{AB} + \mathrm{AC}}{\mathrm{BC}}


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 22 halves, ie,
Ar(
\triangleABD) = Ar(\triangleACD) = Ar(\triangleCAF) = Ar(\triangleCBF) = Ar(\triangleBAE) = Ar(\triangleBCE)

Property 2: The centroid divides each median in the ratio of
2:12 : 1, ie, AK : KD = BK : KE = CK : KF = 2:12 : 1

Property 3: Ar(
\triangleAKE) = Ar(\triangleAKF) = Ar(\triangleBKF) = Ar(\triangleBKD) = Ar(\triangleCKD) = Ar(\triangleCKE) = 16×\dfrac{1}{6} \times Ar(\triangleABC)

Property 4: Ar(
\triangleAFE) = Ar(\triangleCED) = Ar(\triangleBFD) = Ar(\triangleDEF) = 14×\dfrac{1}{4} \times Ar(\triangleABC)


Note: The intersection point of any two medians is the centroid of the triangle. This is because all three medians of any triangle will pass through the same point. The same applies for any two altitudes (which meet at the Orthocentre), any two angle bisectors (which meet at the Incentre) and any two perpendicular bisectors (which meet at the Circumcentre).

Example 10

In \triangle ABC, points D and E are the mid-points of sides BC and CA respectively. The lines AD and BE intersect at point O. What is the ratio of the areas of quadrilateral CDOE and \triangle ABC?

Solution

AD and BE are the medians of \triangle ABC. In the figure below, we draw the third median CF, which will also pass through O, which is the centroid. The 3 medians divide the larger triangle into 6 equal triangles.

Let the area of each smaller triangle be 1 unit.
∴ Area of \triangle ABC == 6 units

Area(CDOE) = Area(
\triangle COD) + Area( \triangle COE) = 2 units

Ratio of areas of CDOE to
\triangle ABC = 2 : 6 = 1 : 3


Answer:
1:31 : 3


Example 11

O is the incentre of \triangle ABC. The line joining A and O intersects BC at D. If \angleADB = 100°100\degree and \angleOCD = 35°35\degree, then ∠BOC = ?

Solution

The incentre is where the angle bisectors meet. Therefore, AO, BO and CO are angle bisectors of \triangle ABC

\angleOCD == \angleOCA = 35°35\degree (CO is the \angle bisector)
\angleADC =180°= 180\degree - \angleADB =80°= 80\degree (Linear Pair)
In
\triangle ADC, \angleDAC + 80°80\degree + 70°70\degree =180°= 180\degree
\angleDAC =30°= 30\degree
\angleBOC =90°= 90\degree + 12\dfrac{1}{2} \angleBAC (Angle bisector Property)
=90°= 90\degree+ \angleDAC =120°= 120\degree


Answer:
120°120\degree


Example 12

In \triangle ABC, D lies on BC such that BD : DC = 1 : 3 and E lies on AC such that AE : EC = 2 : 3. If F lies on DE such that DF : FE = 1 : 2, then what is the ratio of areas of triangles ABC and AFE?

Solution

Let AH be perpendicular to BC. Therefore, AH is the altitude for the bases BC, BD and DC.

Area ( \triangle ABD): Area (\triangle ADC) =12×= \dfrac{1}{2} \times BD ×\times AH : 12×\dfrac{1}{2} \times DC ×\times AH == BD : DC == 1 : 3

Going forward, please note that when a line segment from a vertice to the side divides the side in the ratio of
a:ba : b, the two triangles so formed have areas in the ratio of a:ba : b as the altitudes to the divided sides remains the same..

Lets start with the area of the innermost triangles and use a variable xx.

Area (
\triangle ADF): Area (\triangle AFE) == 1 : 2

∴ Let the actual areas of
\triangle ADF and \triangle AFE be xx and 2xx respectively.

Area of
\triangle ADE =x+2x=3x= x + 2x = 3x

Area (
\triangle ADE) : Area (\triangle CDE) == 2 : 3

3xArea(CDE)=23 \dfrac{3x}{ \mathrm{Area ( \triangle CDE)}} = \dfrac{2}{3}Area(CDE)=4.5x \mathrm{Area ( \triangle CDE)} = 4.5x


Area of
\triangle ADC =3x+4.5x=7.5x= 3x + 4.5x = 7.5x

Area (
\triangle ABD) : Area ( \triangle ADC) == 1 : 3 ⇒ Area(ABD)7.5x=13 \dfrac{ \mathrm{Area ( \triangle ABD)}}{7.5x} = \dfrac{1}{3} ⇒ Area ( \triangle ABD) =2.5x= 2.5x

∴ Area (
\triangle ABC) == Area ( \triangle ABD) + Area ( \triangle ADC) =2.5x+7.5x=10x= 2.5x + 7.5x = 10x

Ratio of areas of triangles ABC and AFE
=10x:2x=5:1= 10x : 2x = 5 : 1

Answer:
5:1 5 : 1


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