CBSE Class 10 Mathematics Triangles Notes Set 10

Important Notes and Formulas

Similar Figures

Two figures are similar when they have exactly the same shape, even if their sizes differ. Every pair of congruent figures is similar, but similar figures need not be congruent. Any two line segments, any two circles, any two squares and any two equilateral triangles are always similar.

Two polygons with the same number of sides are similar only when both of these conditions hold together:
1. Their corresponding angles are equal.
2. Their corresponding sides are in the same ratio.
One condition alone is not enough: a square and a non-square rhombus have proportional sides but different angles, while a square and a non-square rectangle have equal angles but sides that are not proportional.

Similar Triangles

Triangles ABC and DEF are similar, written \( \triangle ABC \sim \triangle DEF \), when ∠A = ∠D, ∠B = ∠E, ∠C = ∠F and \[ \frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}. \] The order of the letters matters, because it records which vertices correspond.

Basic Proportionality Theorem (Thales Theorem)

If a line is drawn parallel to one side of a triangle so that it meets the other two sides at distinct points, it divides those two sides in the same ratio.

In △ABC with D on AB, E on AC and DE ∥ BC:
\[ \frac{AD}{DB}=\frac{AE}{EC} \]Idea of the proof. Join BE and CD. Triangles ADE and DBE share the height from E to AB, so their areas are in the ratio AD : DB. Triangles ADE and DEC share the height from D to AC, so their areas are in the ratio AE : EC. Triangles DBE and DEC lie on the same base DE between the same parallels DE and BC, so they have equal areas. Hence the two ratios are equal.

Equivalent forms that are often useful (each follows by adding or inverting the proportion):
1. \( \dfrac{DB}{AD}=\dfrac{EC}{AE} \)
2. \( \dfrac{AB}{AD}=\dfrac{AC}{AE} \)
3. \( \dfrac{AB}{DB}=\dfrac{AC}{EC} \)
4. \( \dfrac{AD}{AB}=\dfrac{AE}{AC} \)

Converse of the Basic Proportionality Theorem

If a line meets two sides of a triangle and divides them in the same ratio, then it is parallel to the third side. The usual proof is by contradiction: draw through D the line that is really parallel to BC, apply the theorem, and show that it must meet AC at E itself.

Results That Follow From the Theorem

1. The line through the mid-point of one side parallel to a second side bisects the third side.
2. The line joining the mid-points of two sides is parallel to the third side (mid-point theorem) and equals half of it.
3. The diagonals of a trapezium divide each other proportionally, and the converse is also true.
4. A line parallel to the parallel sides of a trapezium divides the non-parallel sides in the same ratio.
5. If three or more parallel lines are cut by two transversals, the intercepts on one transversal are in the same ratio as the corresponding intercepts on the other.
6. Lines drawn parallel to the base through the points that trisect one side also trisect the other side.

The figure shows the mid-point triangle DEF of △ABC. Each of the four small triangles AFE, FBD, EDC and DEF is similar to △ABC, and each has one quarter of its area.

Angle Bisector Theorem

The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the other two sides.

If AD bisects ∠A and meets BC at D, then \[ \frac{BD}{DC}=\frac{AB}{AC}. \] Idea of the proof. Draw CE parallel to DA to meet BA produced at E. Alternate and corresponding angles show that △ACE has equal base angles, so AE = AC. The basic proportionality theorem in △BCE then gives BD/DC = BA/AE = AB/AC.
The converse is also true: if a line from a vertex divides the opposite side in the ratio of the adjacent sides, it bisects the angle at that vertex.
The bisector of an exterior angle at A meets BC produced at a point E that divides BC externally in the same ratio: \( \dfrac{BE}{EC}=\dfrac{AB}{AC} \).

Criteria for Similarity of Triangles

1. AAA (equiangular): if the three angles of one triangle equal the three angles of another, the triangles are similar and their sides are proportional.
2. AA: two equal angles are enough, because the third angles must then be equal too.
3. SSS: if the three pairs of corresponding sides are in the same ratio, the triangles are similar.
4. SAS: if one angle of a triangle equals one angle of another and the sides including these angles are in the same ratio, the triangles are similar.
Two pairs of proportional sides with an angle that is not the included one do not guarantee similarity.

More Properties of Similar Triangles

If △ABC ~ △DEF then the ratio of any pair of corresponding sides is also the ratio of:
1. their corresponding medians,
2. their corresponding altitudes,
3. their corresponding angle bisector segments,
4. their perimeters.
Also, if two sides and the median to one of those sides in a triangle are proportional to the two sides and the corresponding median of another triangle, the triangles are similar; the same holds when the median is drawn to the third side.

Areas of Similar Triangles

The ratio of the areas of two similar triangles equals the ratio of the squares of any two corresponding sides: \[ \frac{ar(\triangle ABC)}{ar(\triangle DEF)}=\frac{AB^2}{DE^2}=\frac{BC^2}{EF^2}=\frac{CA^2}{FD^2}. \] The same ratio is obtained from the squares of corresponding altitudes, medians or angle bisector segments.
Proof idea: area = ½ × base × height, and the heights are in the same ratio as the bases, so the area ratio is (ratio of bases) × (ratio of heights) = (ratio of sides)².
If two similar triangles have equal areas, they are congruent.
A side ratio of k : 1 gives an area ratio of k² : 1, so a 2 : 1 ratio of sides means 4 : 1 in area.

Right Triangle Similarity

If the altitude BD is drawn from the right angle B to the hypotenuse AC, the two small triangles are similar to each other and to the whole triangle: △ADB ~ △BDC ~ △ABC.

This gives three useful relations:
1. \( BD^2 = AD \times DC \)
2. \( AB^2 = AD \times AC \)
3. \( BC^2 = DC \times AC \)

Pythagoras Theorem and Its Converse

In a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides: \( AC^2 = AB^2 + BC^2 \) when ∠B = 90°. A neat proof adds the relations AB² = AD × AC and BC² = DC × AC from the previous section.
Converse: if the square of one side of a triangle equals the sum of the squares of the other two sides, the angle opposite that side is a right angle.
Common Pythagorean triples: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (20, 21, 29).

Extensions of Pythagoras Theorem

1. If ∠B of △ABC is obtuse and AD ⊥ CB (D on CB produced), then \( AC^2 = AB^2 + BC^2 + 2\,BC \cdot BD \).
2. If ∠B is acute and AD ⊥ BC, then \( AC^2 = AB^2 + BC^2 - 2\,BC \cdot BD \).
3. Altitude of an equilateral triangle of side a: \( h = \dfrac{\sqrt3}{2}a \), which also gives 3a² = 4h².
4. Median formula: \( AB^2 + AC^2 = 2AD^2 + 2BD^2 \), where AD is the median to BC.
5. Sum of the squares of the sides: \( 3(AB^2 + BC^2 + CA^2) = 4(AD^2 + BE^2 + CF^2) \), where AD, BE, CF are the medians.
6. In a rhombus, the sum of the squares of the four sides equals the sum of the squares of the diagonals.
7. If CD is the altitude to the hypotenuse of a right triangle with legs a, b and hypotenuse c, then CD · c = ab and \( \dfrac{1}{CD^2}=\dfrac1{a^2}+\dfrac1{b^2} \).

Solved Examples

Example 1. In △ABC, DE ∥ BC with D on AB and E on AC. If AD = 4 cm, DB = (x − 4) cm, AE = 8 cm and EC = (3x − 19) cm, find x.
Solution.
1. By the basic proportionality theorem, \( \dfrac{4}{x-4}=\dfrac{8}{3x-19} \).
2. Cross-multiplying: 4(3x − 19) = 8(x − 4), so 12x − 76 = 8x − 32.
3. 4x = 44 and x = 11.

Example 2. In △PQR, the points A and B lie on PQ and PR with PA = 2 cm, AQ = 6 cm, PB = 3 cm and BR = 9 cm. Is AB ∥ QR?
Solution.
1. PA/AQ = 2/6 = 1/3 and PB/BR = 3/9 = 1/3.
2. The ratios are equal, so by the converse of the basic proportionality theorem AB ∥ QR.

Example 3. In △ABC, AD bisects ∠A and meets BC at D. If AB = 6 cm, AC = 9 cm and BC = 10 cm, find BD and DC.
Solution.
1. BD : DC = AB : AC = 6 : 9 = 2 : 3.
2. BD = 10 × 2/5 = 4 cm and DC = 10 × 3/5 = 6 cm.

Example 4. In △ABC, D lies on AC with ∠ADB = ∠ABC. Show that AB² = AD × AC.
Solution.
1. In △ABD and △ACB, ∠ADB = ∠ABC (given) and ∠A is common.
2. By AA similarity △ABD ~ △ACB, so AB/AC = AD/AB.
3. Hence AB² = AD × AC.

Example 5. The areas of two similar triangles are 81 cm² and 49 cm². If an altitude of the larger triangle is 4.5 cm, find the corresponding altitude of the smaller triangle.
Solution.
1. Ratio of altitudes = square root of the ratio of areas = 9/7.
2. Altitude of the smaller triangle = 4.5 × 7/9 = 3.5 cm.

Example 6. In a trapezium ABCD with AB ∥ DC, the diagonals meet at O and AB = 3 × CD. If ar(△COD) = 12 cm², find ar(△AOB).
Solution.
1. △AOB ~ △COD by AA (alternate angles).
2. Area ratio = (AB/CD)² = 9, so ar(AOB) = 9 × 12 = 108 cm².

Example 7. A pole 6 m high casts a shadow 4 m long. At the same time a building casts a shadow 30 m long. Find the height of the building.
Solution.
1. The sun's rays make the same angle with the ground, so the two right triangles are similar.
2. h/30 = 6/4, so h = 45 m.

Example 8. Two poles of height 12 m and 7 m stand on level ground 12 m apart. Find the distance between their tops.
Solution.
1. The difference in heights is 12 − 7 = 5 m.
2. Distance = √(12² + 5²) = √169 = 13 m.

Example 9. Prove that in a right triangle with legs 6 cm and 8 cm the perpendicular from the right angle to the hypotenuse has length 4.8 cm.
Solution.
1. Hypotenuse = √(36 + 64) = 10 cm.
2. Area in two ways: ½ × 6 × 8 = ½ × 10 × h, so h = 48/10 = 4.8 cm.