CBSE Class 10 Maths Chapter Triangles Notes

The chapter Triangles is one of the most important topics in CBSE Class 10 Mathematics because it introduces students to the concepts of similarity and proportionality in geometric figures. Unlike earlier classes where students mainly studied the basic properties of triangles, this chapter focuses on understanding how triangles can have the same shape but different sizes. Students learn the criteria for similarity of triangles, including the Angle-Angle (AA), Side-Side-Side (SSS), and Side-Angle-Side (SAS) similarity conditions. 

The chapter also explains the relationship between corresponding sides of similar triangles and how these relationships can be used to solve complex geometry problems. Another key concept covered is the Basic Proportionality Theorem and its converse, which help establish proportional relationships within triangles. The chapter concludes with an introduction to the Pythagoras Theorem and its converse, which are widely used in geometry and real-life measurements. Questions from this chapter frequently appear in CBSE board examinations and often require logical reasoning and step-by-step proofs. A strong understanding of triangles not only helps students score well in examinations but also builds a solid foundation for higher-level geometry studied in senior classes. Regular practice of proofs, constructions, and numerical problems is essential for mastering this chapter.Also check out NCERT Exemplar for Class 10 Maths, NCERT Solutions for Class 10, & Maths formulas prepared and developed by experts at Myclass24. 

Comprehensive Class 10 Triangles Notes with Examples

Class 10 Triangles Notes cover all essential theorems, properties, and problem-solving techniques, including congruence, similarity, Pythagoras theorem, and properties of medians, altitudes, and angle bisectors. Each concept is explained with examples and practical applications for easier understanding. These notes help students learn step-by-step methods to solve triangle-related problems efficiently. Understanding triangles is fundamental for geometry, trigonometry, and real-life applications in construction and design. 

CONGRUENT AND SIMILAR FIGURES

Two geometric figures having the same shape and size are known as congruent figures. Geometric figures having the same shape but different sizes are known as similar figures.

SIMILAR TRIANGLES:

Two triangles ABC and DEF are said to be similar if their:

(i) Corresponding angles are equal.

i.e. ∠A = ∠D, ∠B = ∠E, ∠C = ∠F

And,

(ii) Corresponding sides are proportional i.e. AB/DE = BC/EF = AC/DF.

CHARACTERISTIC PROPERTIES OF SIMILAR TRIANGLES:

(i) (AAA Similarity) If two triangles are equiangular, then they are similar.

(ii) (SSS Similarity) If the corresponding sides of two triangles are proportional, then they are similar.

(iii) (SAS Similarity) If in two triangle's one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar.

RESULTS BASED UPON CHARACTERISTIC PROPERTIES OF SIMILAR TRIANGLES:

(i) If two triangles are equiangular, then the ratio of the corresponding sides is the same as the ratio of the corresponding medians.

(ii) If two triangles are equiangular, then the ratio of the corresponding sides is same at the ratio of the corresponding angle bisector segments.

(iii) If two triangles are equiangular then the ratio of the corresponding sides is same at the ratio of the corresponding altitudes.

(iv) If one angle of a triangle is equal to one angle of another triangle and the bisectors of these equal angles divide the opposite side in the same ratio, then the triangles are similar.

(v) If two sides and a median bisecting one of these sides of a triangle are respectively proportional to the two sides and the corresponding median of another triangle, then the triangles are similar.

(vi) If two sides and a median bisecting the third side of a triangle are respectively proportional to the corresponding sides and the median another triangle, then two triangles are similar.

BASIC PROPORTIONALITY THEOREM (THALES THEOREM)

THEOREM 1:

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Given: A triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively.4

To prove: AD/DB = AE/EC

Construction: Join BE and CD and draw DM ⊥ AC and EN ⊥ AB.

Proof: area of ∠ADE

(Taking AD as base)

So, ar(ADE) = 1/2 × AD × EN [The area of ∠ADE is denoted as ar(ADE)].

Similarly, ar(BDE) = 1/2 × DB × EN

and ar(ADE) = 1/2 × AE × DM (Taking AE as base)

Therefore, ar(ADE)/ar(BDE) = AD/DB ...(i)

and ar(ADE)/ar(DEC) = AE/EC ...(ii)

ar(BDE) = ar(DEC) ... (iii)

[∠BDE and DEC are on the same base DE and between the same parallels BC and DE.]

Therefore, from (i), (ii) and (iii), we have: AD/DB = AE/EC.

Corollary: From above equation we have AD/DB = AE/EC.

Adding '1' to both sides we have

AD/DB + 1 = AE/EC + 1

⇒ (AD + DB)/DB = (AE + EC)/EC

⇒ AB/DB = AC/EC.

Theorem 2 - Converse of BPT

THEOREM 2:

(Converse of BPT theorem)

Statement: If a line divides any two sides of a triangle in the same ratio, prove that it is parallel to the third side.

Given: In ∠ABC, DE is a straight line such that AD/DB = AE/EC.

To prove: DE || BC.

Construction: If DE is not parallel to BC, draw DF meeting AC at F.

Proof:

In ∠ABC, let DF || BC

AD/DB = AF/FC ...(i)

[A line drawn parallel to one side of a ∠ divides the other two sides in the same ratio.]

But AD/DB = AE/EC ...(ii) [given]

From (i) and (ii), we get

AF/FC = AE/EC.

Adding 1 to both sides, we get

(AF + FC)/FC = (AE + EC)/EC

AC/FC = AC/EC

FC = EC.

It is possible only when E and F coincide

Hence, DE || BC.

SOME IMPORTANT RESULTS AND THEOREMS:

1. The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.
2. In a triangle ABC, if D is a point on BC such that D divides BC in the ratio AB : AC, then AD is the bisector of ∠A.
3. The external bisector of an angle of a triangle divides the opposite sides externally in the ratio of the sides containing the angle.
4. The line drawn from the mid-point of one side of a triangle parallel to another side bisects the third side.
5. The line joining the mid-points of two sides of a triangle is parallel to the third side.
6. The diagonals of a trapezium divide each other proportionally.
7. If the diagonals of a quadrilateral divide each other proportionally, then it is a trapezium.
8. Any line parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally.
9. If three or more parallel lines are intersected by two transversal, then the intercepts made by them on the transversal are proportional.

SOLVED PROBLEMS

Problem 1:

Question: In a ∠ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 4x – 3, AE = 8x – 7, BD = 3x – 1 and CE = 5x – 3, find the value of x.

Solution:

In ∠ABC, we have

DE||BC.

AD/BD = AE/CE. [By Basic Proportionality Theorem]

(4x - 3)/(3x - 1) = (8x - 7)/(5x - 3)

(4x – 3)(5x – 3) = (8x – 7)(3x – 1)

20x² – 15x – 12x + 9 = 24x² – 21x – 8x + 7

20x² – 27x + 9 = 24x² – 29x + 7

4x² – 2x – 2 = 0

2x² – x – 1 = 0

(2x + 1)(x – 1) = 0

x = 1 or x = –1/2.

So, the required value of x is 1.

[x = -1/2 is neglected as length cannot be negative].

Problem 2:

Question: D and E are respectively the points on the sides AB and AC of a ∠ABC such that AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm, show that DE || BC.

Solution:

We have,

AB = 12 cm, AC = 18 cm, AD = 8 cm and AE = 12 cm.

BD = AB - AD = (12 – 8) cm = 4 cm.

CE = AC – AE = (18 – 12) cm = 6 cm.

Now, AD/BD = 8/4 = 2.

And, AE/CE = 12/6 = 2

AD/BD = AE/CE.

Thus, DE divides sides AB and AC of ∠ABC in the same ratio. Therefore, by the converse of basic proportionality theorem we have DE||BC.

Problem 3:

Question: In a trapezium ABCD AB||DC and DC = 2AB. EF drawn parallel to AB cuts AD in F and BC in E such that BE/EC = 4/3. Diagonal DB intersects EF at G. Prove that 7FE = 10AB.

Solution:

In ∠DFG and ∠DAB,

∠1 = ∠2 [Corresponding ∠s ∵ AB || FG]

∠FDG = ∠ADB [Common]

∠DFG ~ ∠DAB [By AA rule of similarity]

FG/AB = DF/DA .....(i)

Again in trapezium ABCD

EF||AB||DC

BE/EC = AF/FD

4/3 = AF/FD

FD/AF = 3/4

(FD + AF)/AF = (3 + 4)/4

DA/AF = 7/4

AF/DA = 4/7

DF/DA = (DA - AF)/DA = 1 - 4/7 = 3/7 …….(ii)

From (i) and (ii), we get

FG/AB = 3/7 i.e. FG = (3/7)AB ......(iii)

In ∠BEG and ∠BCD, we have

∠BEG = ∠BCD [Corresponding angle ∵ EG||CD]

∠GBE = ∠DBC [Common]

∠BEG ~ ∠BCD [By AA rule of similarity]

EG/CD = BE/BC

EG/CD = BE/(BE + EC)

EG/CD = 4/(4 + 3) = 4/7

EG = (4/7)CD = (4/7) × 2AB = (8/7)AB .....(iv)

Adding (iii) and (iv), we get

FG + EG = (3/7)AB + (8/7)AB

FE = (11/7)AB

7FE = 10AB. Hence proved.

Problem 4:

Question: In ∠ABC, if AD is the bisector of ∠A, prove that AB/AC = BD/DC.

Solution:

In ∠ABC, AD is the bisector of ∠A.

BD/DC = AB/AC ....(i) [By internal bisector theorem]

From A draw AL ⊥ BC

AB/AC = BD/DC. [From (i)] Hence Proved.

Problem 5:

Question: ∠BAC = 90°, AD is its bisector. If DE ⊥ AC, prove that DE × (AB + AC) = AB × AC.

Solution:

It is given that AD is the bisector of ∠A of ∠ABC.

BD/DC = AB/AC

(BD + DC)/DC = (AB + AC)/AC [Adding 1 on both sides]

BC/DC = (AB + AC)/AC

DC/BC = AC/(AB + AC). ...(i)

In ∠'s CDE and CBA, we have

∠DCE = ∠BCA [Common]

∠DEC = ∠BAC. [Each equal to 90°]

So by AA-criterion of similarity

∠CDE ~ ∠CBA

DE/AB = DC/BC ...(ii)

From (i) and (ii), we have

DE/AB = AC/(AB + AC)

DE × (AB + AC) = AB × AC.

CRITERIA FOR SIMILARITY OF TWO TRIANGLES

Two triangles are said to be similar if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportional).

Thus, two triangles ABC and DEF are similar if

(i) ∠A = ∠D, ∠B = ∠E, ∠C = ∠F and

(ii) AB/DE = BC/EF = AC/DF.

In this section, we shall make use of the theorems discussed in earlier sections to derive some criteria for similar triangles which in turn will imply that either of the above two conditions can be used to define the similarity of two triangles.

AAA and SSS Similarity Theorems

CHARACTERISTIC PROPERTY 1 (AAA SIMILARITY)

THEOREM 3:

Statement: If in two triangles, the corresponding angles are equal, then the triangles are similar.

Given: Two triangles ABC and DEF in which ∠A = ∠D, ∠B = ∠E, ∠C = ∠F.

To prove: ∠ABC ~ ∠DEF.

Proof:

Case 1: When AB = DE

In triangles ABC and DEF, we have

∠A = ∠D [Given]

AB = DE [Given]

∠B = ∠E [Given]

∴ ∠ABC ≅ ∠DEF [By ASA congruency]

⇒ BC = EF and AC = DF [c.p.c.t.]

Thus AB/DE = BC/EF = AC/DF [corresponding sides of similar ∠s are proportional]

Hence ∠ABC ~ ∠DEF

Case 2: When AB < DE

Let P and Q be points on DE and DF respectively such that DP = AB and DQ = AC. Join PQ.

In ∠ABC and ∠DPQ, we have

AB = DP [By construction]

∠A = ∠D [Given]

AC = DQ [By construction]

∴ ∠ABC ≅ ∠DPQ [By SAS congruency]

∴ ∠ABC = ∠DPQ [c.p.c.t.] …(i)

But ∠ABC = ∠DEF [Given] …(ii)

∴ ∠DPQ = ∠DEF [c.p.c.t.]

But ∠DPQ and ∠DEF are corresponding angles.

⇒ PQ || EF

∴ DP/DE = DQ/DF [Corollary to BPT Theorem]

∴ AB/DE = AC/DF. [∵ DP = AB and DQ = AC (by construction)]

Similarly AB/DE = BC/EF.

∴ AB/DE = BC/EF = AC/DF.

Hence, ∠ABC ~ ∠DEF.

Case 3: When AB > DE

Let P and Q be points on AB and AC respectively such that AP = DE and AQ = DF. Join PQ.

In ∠APQ and ∠DEF, we have

AP = DE [By construction]

AQ = DF [By construction]

∠A = ∠D [Given]

∴ ∠APQ ≅ ∠DEF [By SAS congruency]

∴ ∠APQ = ∠DEF [c.p.c.t.] …(i)

But ∠DEF = ∠ABC. [Given] …(ii)

From (i) and (ii) we have

∴ ∠APQ = ∠ABC.

But ∠APQ and ∠ABC are corresponding angles

∴ PQ || BC

∴ AP/AB = AQ/AC [Corollary to BPT Theorem]

∴ DE/AB = DF/AC [∵ AP = DE and AQ = DF (by construction)]

Similarly DE/AB = EF/BC.

Thus DE/AB = EF/BC = DF/AC

Or, AB/DE = BC/EF = AC/DF.

Hence ∠ABC ~ ∠DEF.

CHARACTERISTIC PROPERTY 2 (SSS SIMILARITY)

THEOREM 4:

Statement: If the corresponding sides of two triangles are proportional, then they are similar.

Given: Two triangles ABC and DEF such that

AB/DE = BC/EF = CA/FD

To prove: ∠ABC ~ ∠DEF

Construction: Let P and Q be points on DE and DF respectively such that DP = AB and DQ = AC. Join PQ

Proof:

AB/DE = AC/DF [Given]

⇒ DP/DE = DQ/DF …(i)

[∵ DP = AB and DQ = AC (by construction)].

In ∠DEF, we have

DP/DE = DQ/DF [From (i)]

∴ PQ || EF [By the converse of BPT]

∴ ∠DPQ = ∠DEF and ∠DQP = ∠DFE [Corresponding angles]

∴ ∠DPQ ~ ∠DEF [By AA similarity] …(ii)

∴ DP/DE = PQ/EF

or AB/DE = PQ/EF [∵ DP = AB] …(iii)

But AB/DE = BC/EF [given] …(iv)

From equations (iii) and (iv), we have

PQ/EF = BC/EF

⇒ BC = PQ …(v)

In ∠ABC and ∠DPQ, we have

AB = DP [By construction]

AC = DQ [By construction]

BC = PQ [By (v)]

∴ ∠ABC ≅ ∠DPQ [by SSS congruency]

⇒ ∠ABC ~ ∠DPQ. [∵ ∠ABC ≅ ∠DPQ ⇔ ∠ABC ~ ∠DPQ] ... (vi)

From equation (ii) and (vi), we get

∠ABC ~ ∠DEF.

CHARACTERISTIC PROPERTY 3 (SAS SIMILARITY)

THEOREM 5:

Statement: If one angle of a triangle is equal to one angle of the other and the sides including these angles are proportional than the two triangles are similar.

Given: Two triangle ABC and DEF such that ∠A = ∠D and AB/DE = AC/DF.

To prove: ∠ABC ~ ∠DEF.

Construction: Let P and Q be points on DE and DF respectively such that DP = AB and DQ = AC. Join PQ.

Proof:

In ∠ABC and ∠DPQ, we have

AB = DP [By construction]

AC = DQ [By construction]

∠A = ∠D [Given]

∠ABC ≅ ∠DPQ [By SAS congruency] …(i)

Now, AB/DE = AC/DF …(ii)

DP/DE = DQ/DF. [∵ AB = DP and AC = DQ (by construction)]

In ∠DEF, we have

DP/DE = DQ/DF. [From (ii)]

PQ || EF [By the converse of BPT]

∠DPQ = ∠DEF and ∠DQP = ∠DFE [Corresponding angles]

∠DPQ ~ ∠DEF [By AA similarity] …(iii)

From equations (i) and (iii), we get ∠ABC ~ ∠DEF.

SOLVED PROBLEMS

PROPERTIES OF AREAS OF SIMILAR TRIANGLES:

1. The areas of two similar triangles are in the ratio of the squares of corresponding altitudes.
2. The areas of two similar triangles are in the ratio of the squares of the corresponding medians.
3. The area of two similar triangles are in the ratio of the squares of the corresponding angle bisector segments.

SOLVED PROBLEMS ON AREAS

Problem 1:

Question: Prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on this diagonals.

Solution:

Given: A square ABCD. Equilateral triangles BCE and ACF have been described on side BC and diagonals AC respectively.

To prove: Area (BCE) = (1/2) Area (ACF).

Proof:

Since ∠BCE and ∠ACF are equilateral. Therefore, they are equiangular (each angle being equal to 60°) and hence ∠BCE ~ ∠ACF.

ar(BCE)/ar(ACF) = BC²/AC²

= BC²/(√2 BC)² [∵ AC = √2 BC]

= BC²/(2BC²) = 1/2

ar(BCE) = (1/2) ar(ACF).

Problem 2:

Question: The areas of two similar triangles ABC and PQR are 64 cm² and 36 cm² respectively. If QR = 16.5 cm, find BC.

Solution:

Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.

ar(ABC)/ar(PQR) = BC²/QR²

64/36 = BC²/(16.5)²

BC²/272.25 = 16/9

BC² = (16 × 272.25)/9 = 484

BC = √484 = 22

Hence BC = 22 cm.

PYTHAGORAS THEOREM

Statement:

In a right triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides.

Given: A right triangle ABC, right angled at B.

To prove: AC² = AB² + BC².

Construction: Draw BD ⊥ AC.

Proof:

In ∠ADB and ∠ABC, we have

∠ADB = ∠ABC [Each equal to 90°]

∠A = ∠A [Common]

∠ADB ~ ∠ABC [By AA similarity]

AD/AB = AB/AC [Corresponding sides of similar triangles are proportional]

AB² = AC × AD …(i)

In ∠BCD and ∠ACB, we have

∠CDB = ∠CBA [Each equal to 90°]

∠C = ∠C. [Common]

By AA similarity criterion

∠BCD ~ ∠ACB

CD/BC = BC/AC

BC² = AC × CD …(ii)

Adding equations (i) and (ii), we get

AB² + BC² = AC × AD + AC × CD

AB² + BC² = AC(AD + CD)

AB² + BC² = AC × AC

Hence, AC² = AB² + BC². Hence Proved.

CONVERSE OF PYTHAGORAS THEOREM

Statement:

In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle.

Given: A triangle ABC such that AC² = AB² + BC².

Construction: Construct a triangle DEF such that DE = AB, EF = BC and ∠E = 90°.

Proof:

In order to prove that ∠B = 90°, it is sufficient to show ∠ABC ≅ ∠DEF. For this we proceed as follows Since ∠DEF is a right – angled triangle with right angle at E. Therefore, by Pythagoras theorem, we have

DF² = DE² + EF²

DF² = AB² + BC² [∵ DE = AB and EF = BC (By construction)]

DF² = AC² [∵ AB² + BC² = AC² (Given)]

DF = AC .....(i)

Thus, in ∠ABC and ∠DEF, we have

AB = DE, BC = EF. [By construction]

And AC = DF [From equation (i)]

∠ABC ≅ ∠DEF [By SSS criteria of congruency]

∠B = ∠E = 90°.

Hence, ∠ABC is a right triangle, right angled at B.

SOLVED PROBLEMS

Problem 1:

Question: In a ∠ABC, AB = BC = CA = 2a and AD ⊥ BC. Prove that

(i) AD = a√3.

(ii) area (∠ABC) = a²√3.

Solution:

(i) Here, AD ⊥ BC.

Clearly, ∠ABC is an equilateral triangle.

Thus, in ∠ABD and ∠ACD

AD = AD [Common]

∠ADB = ∠ADC. [90° each]

And AB = AC.

by RHS congruency condition

∠ABD ≅ ∠ACD

BD = DC = a.

Now, ∠ABD is a right angled triangle

AD = √(AB² - BD²) [Using Pythagoras Theorem]

AD = √((2a)² - a²)

AD = √(4a² - a²)

AD = √(3a²)

AD = a√3.

(ii) Area (∠ABC) = (1/2) × BC × AD

= (1/2) × 2a × a√3

= a²√3.

Problem 2:

Question: BL and CM are medians of ∠ABC right angled at A. Prove that 4(BL² + CM²) = 5 BC².

Solution:

In ∠BAL

BL² = AL² + AB² ....(i) [Using Pythagoras theorem]

and In ∠CAM

CM² = AM² + AC² .....(ii) [Using Pythagoras theorem]

Adding (i) and (ii) and then multiplying by 4, we get

4(BL² + CM²) = 4(AL² + AB² + AM² + AC²)

= 4{AL² + AM² + (AB² + AC²)} [∵ ∠ABC is a right triangle]

= 4(AL² + AM² + BC²)

= 4(ML² + BC²) [∵ ∠LAM is a right triangle]

= 4ML² + 4 BC²

= 4(BC/2)² + 4BC²

[A line joining mid-points of two sides is parallel to third side and is equal to half of it, ML = BC/2]

= 4(BC²/4) + 4BC²

= BC² + 4BC² = 5BC². Hence proved.

PRACTICE QUESTIONS - SET 3

11. A ladder is placed in such a way that its foot is at a distance of 5 m from a wall and its top reaches a window 12 m above the ground. Determine the length of the ladder.

12. A ladder 15 m long reaches a window which is 9 m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 12 m high. Find the width of the street.

13. In the given figure, ∠ACB = 90° and CD ⊥ AB. Prove that BC²/AC² = BD/AD.

14. ABC is a triangle right-angled at C and p is the length of the perpendicular from C to AB. Show that (a) pc = ab. (b) 1/p² = 1/a² + 1/b² where a = BC, b = AC and c = AB.

15. In ∠ABC, ∠C > 90° and side AC is produced to D such that segment BD is perpendicular to segment AD. Prove that AB² = AC² + BC² + 2AC.CD

DETAILED SOLUTIONS - SET 3

Question: A ladder is placed in such a way that its foot is at a distance of 5 m from a wall and its top reaches a window 12 m above the ground. Determine the length of the ladder.

Solution:

Let AB be the ladder, B be the window and BC be the wall.

Then BC = 12 m, AC = 5m and ∠ACB = 90°.

In right triangle ACB, we have

AB² = BC² + AC² [By Pythagoras Theorem]

AB² = (12)² + (5)²

AB² = 144 + 25 = 169

AB = √169 = 13 m.

Hence the length of the ladder is 13 m.

Question: A ladder 15 m long reaches a window which is 9 m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 12 m high. Find the width of the street.

Solution:

Let AB be the street and let C and D be the windows at heights of 9 m and 12 m respectively from the ground.

Let E be the foot of the ladder. Then EC and ED are the two positions of the ladder.

Clearly AC = 9 m, BD = 12 m, EC = ED = 15 m and ∠CAE = ∠DBE = 90°

In right triangle CAE, we have

EC² = AC² + AE² [By Pythagoras Theorem]

(15)² = (9)² + AE²

AE² = (15)² – (9)² = (225 – 81) = 144

AE = 12 m. …(i)

In right triangle DBE, we have

ED² = BD² + EB² [By Pythagoras Theorem]

(15)² = (12)² + EB²

EB² = (15)² – (12)² = (225 – 144) = 81

EB = 9 m …(ii)

Adding equations (i) and (ii), we get

AE + EB = (12 + 9) m ⇒ AB = 21 m

Hence, the width of the street is 21 m.

Question: In the given figure, ∠ACB = 90° and CD ⊥ AB. Prove that BC²/AC² = BD/AD.

Solution:

Given: ∠ACB = 90° and CD ⊥ AB.

To prove: BC²/AC² = BD/AD.

Proof:

In ∠ACD and ∠ABC

∠A = ∠A [Common]

∠ADC = ∠ACB [Both 90°]

∠ACD ~ ∠ABC [By AA similarity]

So AC/AB = AD/AC

or, AC² = AB × AD. …(i)

Similarly, ∠BCD ~ ∠BAC.

So BC/AB = BD/BC

or, BC² = AB × BD. …(ii)

Therefore, from (i) and (ii),

BC²/AC² = (AB × BD)/(AB × AD) = BD/AD.

Question: ABC is a triangle right-angled at C and p is the length of the perpendicular from C to AB. Show that (a) pc = ab. (b) 1/p² = 1/a² + 1/b² where a = BC, b = AC and c = AB.

Solution:

(a) Taking c as the base and p as the altitude, we have

area of ∠ABC = (1/2) × c × p …(i)

Taking b as the base and a as the altitude, we have

area ∠ABC = (1/2) × a × b …(ii)

∴ (1/2) × c × p = (1/2) × a × b [From (i) and (ii)]

⇒ pc = ab Hence Proved.

(b) ∠ABC is a right triangle-angled at C.

∴ c² = a² + b² …(iii)

[By Pythagoras Theorem]

pc = ab [proved above]

or p = ab/c

∴ p² = (ab)²/c² [From equation (iii)]

p² = (ab)²/(a² + b²)

1/p² = (a² + b²)/(ab)²

1/p² = (a² + b²)/(a²b²)

1/p² = a²/(a²b²) + b²/(a²b²)

1/p² = 1/b² + 1/a².