11.1 Congruent Triangles, Bisectors & the Projection Rule

Congruent Triangles, Bisectors & the Projection Rule

The NTS Geometry block lists ten named topics, and four of them concern triangles directly: Congruent Triangles, Parallelograms and Triangles, Line Bisectors and Angle Bisectors, and Projection of a side of a triangle. These are the theorem-based topics from Pakistani intermediate geometry, and they are asked as reasoning items rather than as calculations.

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Congruence: Same Shape, Same Size

Two triangles are congruent when every pair of corresponding sides and angles is equal. You never need all six facts to prove it — four criteria suffice, plus one for right triangles.

The Two Traps

AAA is not a congruence criterion. Three equal angles fix the shape but not the size, so AAA proves only similarity. Any option offering AAA as proof of congruence is wrong by definition.

SSA is not a criterion either (except in the RHS special case). Two sides and a non-included angle can produce two different triangles — the ambiguous case. Note that SAS requires the angle to lie between the two sides.

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Similarity: Same Shape, Any Size

Triangles are similar when corresponding angles are equal and corresponding sides are proportional. Three criteria establish it: AA (two angles suffice, since the third follows), SSS proportionality, and SAS proportionality.

The Area Rule

If corresponding sides are in ratio $k$, then:

$\frac{\text{Perimeter}_1}{\text{Perimeter}_2} = k \qquad\qquad \frac{\text{Area}_1}{\text{Area}_2} = k^{2}$

So triangles with sides in ratio $3:5$ have areas in ratio $9:25$. Answering $3:5$ for the area ratio is the single most common similarity error on the paper.

Worked example. Two similar triangles have areas 48 cm² and 108 cm². The smaller has a base of 8 cm. Find the corresponding base of the larger.

$k^{2} = \frac{108}{48} = \frac{9}{4} \;\Rightarrow\; k = \frac{3}{2} \;\Rightarrow\; \text{base} = 8 \times \frac{3}{2} = 12\text{ cm}$

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Line Bisectors and Angle Bisectors

Perpendicular Bisector of a Segment

Every point on the perpendicular bisector of a segment is equidistant from its two endpoints, and the converse holds. The three perpendicular bisectors of a triangle's sides meet at the circumcentre, the centre of the circle through all three vertices.

The Angle Bisector Theorem

The bisector of an interior angle divides the opposite side in the ratio of the two adjacent sides:

$\frac{BD}{DC} = \frac{AB}{AC}$

where $AD$ bisects angle $A$ and meets $BC$ at $D$.

Worked example. In triangle $ABC$, $AB = 12$, $AC = 18$, and $BC = 20$. The bisector from $A$ meets $BC$ at $D$. Find $BD$.

$\frac{BD}{DC} = \frac{12}{18} = \frac{2}{3} \;\Rightarrow\; BD = \frac{2}{5}\times 20 = 8, \qquad DC = 12$

Every point on an angle bisector is equidistant from the two arms of the angle. The three interior bisectors meet at the incentre, the centre of the inscribed circle.

The Four Centres

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Parallelograms and Triangles

• Opposite sides and opposite angles of a parallelogram are equal; consecutive angles are supplementary.
• The diagonals bisect each other, and each diagonal splits the parallelogram into two congruent triangles.
• A triangle and a parallelogram on the same base and between the same parallels satisfy: the triangle's area is half the parallelogram's.
• The segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length (the midpoint theorem).

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Projection of a Side of a Triangle

This theorem generalises Pythagoras to triangles that are not right-angled, and it appears in the NTS list by name.

Drop a perpendicular from $A$ to line $BC$, meeting it at $D$. The segment $BD$ is the projection of $AB$ onto $BC$.

Acute case. When the angle opposite side $a$ is acute:

$a^{2} = b^{2} + c^{2} - 2\,b \cdot (\text{projection of } c \text{ on } b)$

Obtuse case. When that angle is obtuse, the projection falls outside the triangle and the sign reverses:

$a^{2} = b^{2} + c^{2} + 2\,b \cdot (\text{projection of } c \text{ on } b)$

Because the projection equals $c\cos A$, both statements are the law of cosines in geometric dress:

$a^{2} = b^{2} + c^{2} - 2bc\cos A$

When $A = 90°$, $\cos A = 0$ and the formula reduces to Pythagoras. There is also a companion projection formula expressing a side as the sum of two projections:

$a = b\cos C + c\cos B$

Worked example. A triangle has $b = 7$, $c = 5$, and the included angle $A = 60°$.

$a^{2} = 49 + 25 - 2(7)(5)\cos 60° = 74 - 70(0.5) = 74 - 35 = 39 \;\Rightarrow\; a = \sqrt{39} \approx 6.24$

Diagnostic worth memorising: compare $a^{2}$ with $b^{2} + c^{2}$. If $a^{2} < b^{2}+c^{2}$ the angle opposite $a$ is acute; if equal, right; if greater, obtuse. Sides 6, 8, 11 give $121 > 100$, so the triangle is obtuse — and NTS asks this without requiring a single trigonometric value.