FBISE, NCP & NBF-aligned resources for Classes 9–12
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Math Chapter 10 : Topic 1
Triangle Lines and Concurrency
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In a triangle, an angle bisector is a line segment that:
Reasoning: An angle bisector divides an angle of the triangle into two angles of equal measure.
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The three internal angle bisectors of a triangle meet at a point called the:
Reasoning: The common point of the three internal angle bisectors is called the incenter.
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Which property can be used to verify that a point is the incenter of a triangle?
Reasoning: The incenter is equidistant from the three sides of a triangle.
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While constructing the angle bisector of $\angle A$, an arc centred at $A$ cuts $AB$ and $AC$ at $P$ and $Q$. The next step is to:
Reasoning: Equal arcs from $P$ and $Q$ intersect at a point which, when joined to $A$, gives the angle bisector.
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The incenter of every triangle always lies:
Reasoning: The three internal angle bisectors always intersect inside the triangle.
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A perpendicular bisector of a side of a triangle:
Reasoning: A perpendicular bisector crosses a segment at its midpoint and forms a right angle with it.
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The perpendicular bisectors of the sides of a triangle are concurrent at the:
Reasoning: The common point of the perpendicular bisectors is the circumcenter.
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A point $O$ lies on the perpendicular bisector of $AB$. Which relation must be true?
Reasoning: Every point on the perpendicular bisector of a segment is equidistant from its endpoints.
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If $O$ is the circumcenter of $\mathrm{\Delta}ABC$, then:
Reasoning: The circumcenter is equally distant from all three vertices of a triangle.
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To construct the perpendicular bisector of side $AB$, equal arcs are drawn from $A$ and $B$. The compass opening should be:
Reasoning: The radius must be greater than half of $AB$ so that the arcs intersect at two points.
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Where is the circumcenter located in an acute-angled triangle?
Reasoning: In an acute triangle, all perpendicular bisectors meet inside the triangle.
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In a right-angled triangle, the circumcenter lies at the:
Reasoning: The midpoint of the hypotenuse is equidistant from all three vertices of a right triangle.
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The circumcenter of an obtuse-angled triangle lies:
Reasoning: In an obtuse triangle, the perpendicular bisectors intersect outside the triangle.
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A median of a triangle is a line segment joining:
Reasoning: A median connects a vertex with the midpoint of the side opposite that vertex.
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The three medians of a triangle meet at the:
Reasoning: The point of concurrency of the three medians is called the centroid.
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The centroid divides each median in the ratio:
Reasoning: The distance from the vertex to the centroid is twice the distance from the centroid to the midpoint.
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In triangle $ABC$, $AD$ is a median and $G$ is the centroid. If $AD = 15\ cm$, then $AG$ is:
Reasoning: The centroid divides each median in the ratio $AG\ :GD = 2\ :1$.Therefore, $AG$ is two-thirds of the complete median $AD$.$AD =$ $\frac{2}{3} \times AD$ $\Longrightarrow$ $AD =$ $\frac{2}{3} \times 15$ $\Longrightarrow$ $AD = 10\ cm$
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A student draws two medians of a triangle and marks their point of intersection as $G$. To verify the concurrency of all medians, the student should:
Reasoning: If the third median passes through the intersection of the first two medians, their concurrency is verified.
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An altitude of a triangle is a perpendicular segment drawn from:
Reasoning: An altitude starts at a vertex and meets the opposite side, or its extension, at $90^{o}$.
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The three altitudes of a triangle are concurrent at the:
Reasoning: The common point of the three altitudes is called the orthocenter.
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In a right-angled triangle, two of its altitudes are:
Reasoning: The two legs are perpendicular to each other, so each acts as an altitude from the opposite vertex.
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The orthocenter of a right-angled triangle is located:
Reasoning: Since the two legs themselves are altitudes, they intersect at the right-angle vertex.
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For an obtuse-angled triangle, some altitudes must be drawn to:
Reasoning: From the acute-angle vertices, perpendiculars meet the extensions of the opposite sides.
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Where is the orthocenter of an obtuse-angled triangle located?
Reasoning: The extended altitudes of an obtuse triangle intersect outside the triangle.
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In an acute-angled triangle, the orthocenter lies:
Reasoning: In an acute triangle, all three altitudes meet inside the triangle.
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In an isosceles triangle $ABC$, where $AB = AC$, a line is drawn from $A$ to the midpoint of $BC$. This line is simultaneously:
Reasoning: In an isosceles triangle, the line from the vertex between the equal sides to the base midpoint has all four properties.
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In an equilateral triangle, which statement is correct?
Reasoning: Due to complete symmetry, all medians, altitudes, angle bisectors, and perpendicular bisectors meet at one point.
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A line from vertex $A$ meets side $BC$ at $D$, and $BD = DC$. Which conclusion is certainly correct?
Reasoning: Since $D$ is the midpoint of $BC$, the segment from $A$ to $D$ is a median.
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A line from vertex $A$ meets side $BC$ at point $D$ and $\angle ADB = 90^{o}$. Which conclusion is certainly correct?
Reasoning: A perpendicular segment drawn from a vertex to the opposite side is an altitude.
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A point $P$ inside a triangle is equally distant from all three sides. Which three lines must pass through $P$?
Reasoning: A point equidistant from the sides lies on the angle bisectors, so it is the incenter.
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A point $Q$ is equally distant from vertices $A$, $B$ and $C$. Which lines are concurrent at $Q$?
Reasoning: The point equidistant from all vertices is the circumcenter, formed by the perpendicular bisectors.
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Which line must pass through a vertex of a triangle?
Reasoning: A median is defined as a segment drawn from a vertex to the midpoint of the opposite side.
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Which construction does not necessarily pass through a vertex?
Reasoning: A perpendicular bisector passes through the midpoint of a side and generally does not pass through a vertex.
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In triangle $ABC$, two perpendicular bisectors meet at $O$. Without drawing the third perpendicular bisector, we can conclude that:
Reasoning: All three perpendicular bisectors of a triangle are concurrent at the circumcenter.
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A student incorrectly claims that every median is also an altitude. Which triangle provides the clearest counterexample?
Reasoning: In a general scalene triangle, a median usually does not meet the opposite side at $90^{o}$.
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A triangle is drawn on cardboard. The point at which it can balance horizontally is generally its:
Reasoning: For a uniform triangular sheet, the centroid represents its centre of gravity.
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