**Pythagoras Theorem**

**Theorem 1:** In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

**Given:** A right-angled triangle ABC in which B = ∠90º.

**To Prove:** (Hypotenuse)^{2} = (Base)^{2} + (Perpendicular)^{2}.

i.e., AC^{2} = AB^{2} + BC^{2}

**Construction:** From B draw BD ⊥ AC.

**Proof:** In triangle ADB and ABC, we have

∠ADB = ∠ABC [Each equal to 90º]

and, ∠A = ∠A [Common]

So, by AA-similarity criterion, we have

∆ADB ~ ∆ABC

\(\Rightarrow \frac{AD}{AB}=\frac{AB}{AC}\) [∵ In similar triangles corresponding sides are proportional]

⇒ AB^{2} = AD × AC ….(i)

In triangles BDC and ABC, we have

∠CDB = ∠ABC [Each equal to 90º]

and, ∠C = ∠C [Common]

So, by AA-similarity criterion, we have

∆BDC ~ ∆ABC

\(\Rightarrow \frac{DC}{BC}=\frac{BC}{AC}\) [∵ In similar triangles corresponding sides are proportional]

⇒ BC^{2} = AC × DC ….(ii)

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

AB^{2} + BC^{2} = AD × AC + AC × DC

⇒ AB^{2} + BC^{2} = AC (AD + DC)

⇒ AB^{2} + BC^{2} = AC × AC

⇒ AC^{2} = AB^{2} + BC^{2}

Hence, AC^{2} = AB^{2} + BC^{2}

The converse of the above theorem is also true as proved below.

**Read also:**

**Theorem 2:** (Converse of Pythagoras Theorem).

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 side is a right angle.

**Given:** A triangle ABC such that AC^{2} = AB^{2} + BC^{2}

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

**Proof:** In order to prove that B = ∠90º, it is sufficient to show that ∆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

DF^{2} = DE^{2} + EF^{2}

⇒ DF^{2} = AB^{2} + BC^{2} [∵ DE = AB and EF = BC (By construction)]

⇒ DF^{2} = AC^{2 } [∵ AB^{2} + BC^{2} = AC^{2} (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

⇒ ∠B = ∠E = 90º

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

**Pythagoras Theorem With Examples**

**Example 1:** Side of a triangle is given, determine it is a right triangle.

(2a – 1) cm, \(2\sqrt { 2a } \) cm, and (2a + 1) cm

**Sol.** Let p = (2a – 1) cm, q = \(2\sqrt { 2a } \) cm and r = (2a + 1) cm.

Then, (p^{2} + q^{2}) = (2a – 1)^{2} cm^{2} + (2 )^{2} cm^{2}

= {(4a^{2} + 1– 4a) + 8a}cm^{2}

= (4a^{2} + 4a + 1)cm^{2}

= (2a + 1)^{2} cm^{2} = r^{2}.

(p^{2} + q^{2}) = r^{2}.

Hence, the given triangle is right angled.

**Example 2:** A man goes 10 m due east and then 24 m due north. Find the distance from the starting point.

**Sol. **Let the initial position of the man be O and his final position be B. Since the man goes 10 m due east and then 24 m due north. Therefore, ∆AOB is a right triangle right-angled at A such that OA = 10 m and AB = 24 m.

By Phythagoras theorem, we have

OB^{2} = OA^{2} + AB^{2}

⇒ OB^{2} = 10^{2} + 24^{2} = 100 + 576 = 676

⇒ OB = \(\sqrt { 676 } \) = 26 m

Hence, the man is at a distance of 26 m from the starting point.

**Example 3:** Two towers of heights 10 m and 30 m stand on a plane ground. If the distance between their feet is 15 m, find the distance between their tops.

**Sol.**

By Phythagoras theorem, we have

AC^{2} = CE^{2} + AE^{2}

⇒ AC^{2} = 15^{2} + 20^{2} = 225 + 400 = 625

⇒ AC = \(\sqrt { 625 } \) = 25 m.

**Example 4:** In Fig., ∆ABC is an obtuse triangle, obtuse angled at B. If AD ⊥ CB, prove that

AC^{2} = AB^{2} + BC^{2} + 2BC × BD

**Sol. Given:** An obtuse triangle ABC, obtuse-angled at B and AD is perpendicular to CB produced.

**To Prove:** AC^{2} = AB^{2} + BC^{2} + 2BC × BD

**Proof:** Since ∆ADB is a right triangle right angled at D. Therefore, by Pythagoras theorem, we have AB^{2} = AD^{2} + DB^{2} ….(i)

Again ∆ADC is a right triangle right angled at D.

Therefore, by Phythagoras theorem, we have

AC^{2} = AD^{2} + DC^{2}

⇒ AC^{2} = AD^{2} + (DB + BC)^{2}

⇒ AC^{2} = AD^{2} + DB^{2} + BC^{2} + 2BC • BD

⇒ AC^{2} = AB^{2} + BC^{2} + 2BC • BD [Using (i)]

Hence, AC^{2} = AB^{2} + BC^{2} + 2BC • BD

**Example 5:** In figure, ∠B of ∆ABC is an acute angle and AD ⊥ BC, prove that

AC^{2} = AB^{2} + BC^{2} – 2BC × BD

**Sol. Given:** A ∆ABC in which ∠B is an acute angle and AD ⊥ BC.

**To Prove:** AC^{2} = AB^{2} + BC^{2} – 2BC × BD.

**Proof:** Since ∆ADB is a right triangle right-angled at D. So, by Pythagoras theorem, we have

AB^{2} = AD^{2} + BD^{2} ….(i)

Again ∆ADC is a right triangle right angled at D.

So, by Pythagoras theorem, we have

AC^{2} = AD^{2} + DC^{2}

⇒ AC^{2} = AD^{2} + (BC – BD)^{2}

⇒ AC^{2} = AD^{2} + (BC^{2} + BD^{2} – 2BC • BD)

⇒ AC^{2} = (AD^{2} + BD^{2}) + BC^{2} – 2BC • BD

⇒ AC^{2} = AB^{2} + BC^{2} – 2BC • BD [Using (i)]

Hence, AC^{2} = AB^{2} + BC^{2} – 2BC • BD

**Example 6:** If ABC is an equilateral triangle of side a, prove that its altitude = \(\frac { \sqrt { 3 } }{ 2 } a\).

**Sol. **∆ABD is an equilateral triangle.

We are given that AB = BC = CA = a.

AD is the altitude, i.e., AD ⊥ BC.

Now, in right angled triangles ABD and ACD, we have

AB = AC (Given)

and AD = AD (Common side)

∆ABD ≅ ∆ACD (By RHS congruence)

⇒ BD = CD ⇒ BD = DC = \(\frac { 1 }{ 2 }BC\) = \(\frac { a }{ 2 }\)

From right triangle ABD.

AB^{2} = AD^{2} + BD^{2}

\(\Rightarrow {{a}^{2}}=A{{D}^{2}}+{{\left( \frac{a}{2} \right)}^{2}} \)

\(\Rightarrow A{{D}^{2}}={{a}^{2}}-\frac{{{a}^{2}}}{4}=\frac{3}{4}{{a}^{2}}\)

\(\Rightarrow AD=\frac{\sqrt{3}}{2}a \)

**Example 7:** ABC is a right-angled triangle, right-angled at A. A circle is inscribed in it. The lengths of the two sides containing the right angle are 5 cm and 12 cm. Find the radius of the circle.

**Sol. **Given that ∆ABC is right angled at A.

AC = 5 cm and AB = 12 cm

BC^{2} = AC^{2} + AB^{2} = 25 + 144 = 169

BC = 13 cm

Join OA, OB, OC

Let the radius of the inscribed circle be r

Area of ∆ABC = Area of ∆OAB + Area of ∆OBC + Area of ∆OCA

⇒ 1/2 × AB × AC

\(=\frac{1}{2}\left( 12\text{ }\times \text{ }r \right)\text{ }+\frac{1}{2}\left( 13\text{ }\times \text{ }r \right)\text{ }+\frac{1}{2}\left( 5\text{ }\times \text{ }r \right)\)

⇒ 12 × 5 = r × {12 + 13 + 5}

⇒ 60 = r × 30 ⇒ r = 2 cm

**Example 7:** ABCD is a rhombus. Prove that

AB^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2} + BD^{2}

**Sol. **Let the diagonals AC and BD of rhombus ABCD intersect at O.

Since the diagonals of a rhombus bisect each other at right angles.

∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = 90º

and AO = CO, BO = OD.

Since ∆AOB is a right triangle right-angle at O.

∴ AB^{2} = OA^{2} + OB^{2}

\(\Rightarrow A{{B}^{2}}={{\left( \frac{1}{2}AC \right)}^{2}}+{{\left( \frac{1}{2}BD \right)}^{2}}\) [∵ OA = OC and OB = OD]

⇒ 4AB^{2} = AC^{2} + BD^{2} ….(i)

Similarly, we have

4BC^{2} = AC^{2} + BD^{2} ….(ii)

4CD^{2} = AC^{2} + BD^{2} ….(iii)

and, 4AD^{2} = AC^{2} + BD^{2} ….(iv)

Adding all these results, we get

4(AB^{2} + BC^{2} + AD^{2}) = 4(AC^{2} + BD^{2})

⇒ AB^{2} + BC^{2} + AD^{2} + DA^{2} = AC^{2} + BD^{2}

**Example 8:** P and Q are the mid-points of the sides CA and CB respectively of a ∆ABC, right angled at C. Prove that:

(i) 4AQ^{2} = 4AC^{2} + BC^{2}

(ii) 4BP^{2} = 4BC^{2} + AC^{2}

(iii) (4AQ^{2} + BP^{2}) = 5AB^{2}

**Sol. **

**(i) **Since ∆AQC is a right triangle right-angled at C.

∴ AQ^{2} = AC^{2} + QC^{2}

⇒ 4AQ^{2} = 4AC^{2} + 4QC^{2 }[Multiplying both sides by 4]

⇒ 4AQ^{2} = 4AC^{2} + (2QC)^{2}

⇒ 4AQ^{2} = 4AC^{2} + BC^{2} [∵ BC = 2QC]

**(ii) **Since ∆BPC is a right triangle right-angled at C.

∴ BP^{2} = BC^{2} + CP^{2}

⇒ 4BP^{2} = 4BC^{2} + 4CP^{2 }[Multiplying both sides by 4]

⇒ 4BP^{2} = 4BC^{2} + (2CP)^{2}

⇒ 4BP^{2} = 4BC^{2} + AC^{2} [∵ AC = 2CP]

**(iii) **From (i) and (ii), we have

4AQ^{2} = 4AC^{2} + BC^{2} and, 4BC^{2} = 4BC^{2} + AC^{2}

∴ 4AQ^{2} + 4BP^{2} = (4AC^{2} + BC^{2}) + (4BC^{2} + AC^{2})

⇒ 4(AQ^{2} + BP^{2}) = 5 (AC^{2} + BC^{2})

⇒ 4(AQ^{2} + BP^{2}) = 5 AB^{2}

[In ∆ABC, we have AB^{2} = AC^{2} + BC^{2}]

**Example 9:** From a point O in the interior of a ∆ABC, perpendicular OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that :

(i) AF^{2} + BD^{2} + CE^{2} = OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2}

(ii) AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}

**Sol. **

Let O be a point in the interior of ∆ABC and let OD ⊥ BC, OE ⊥ CA and OF ⊥ AB.

**(i)** In right triangles ∆OFA, ∆ODB and ∆OEC, we have

OA^{2} = AF^{2} + OF^{2}

OB^{2} = BD^{2} + OD^{2}

and, OC^{2} = CE^{2} + OE^{2}

Adding all these results, we get

OA^{2} + OB^{2} + OC^{2} = AF^{2} + BD^{2} + CE^{2} + OF^{2} + OD^{2} + OE^{2}

⇒ AF^{2} + BD^{2} + CE^{2} = OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2}

**(ii)** In right triangles ∆ODB and ∆ODC, we have

OB^{2} = OD^{2} + BD^{2}

and, OC^{2} = OD^{2} + CD^{2}

OB^{2} – OC^{2} = (OD^{2} + BD^{2}) – (OD^{2} + CD^{2})

⇒ OB^{2} – OC^{2} = BD^{2} – CD^{2 } ….(i)

Similarity, we have

OC^{2} – OA^{2} = CE^{2} – AE^{2} ….(ii)

and, OA^{2} – OB^{2} = AF^{2} – BF^{2} ….(iii)

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

(OB^{2} – OC^{2}) + (OC^{2} – OA^{2}) + (OA^{2} – OB^{2})

= (BD^{2} – CD^{2}) + (CE^{2} – AE^{2}) + (AF^{2} – BF^{2})

⇒ (BD^{2} + CE^{2} + AF^{2}) – (AE^{2} + CD^{2} + BF^{2}) = 0

⇒ AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}

**Example 10:** In a right triangle ABC right-angled at C, P and Q are the points on the sides CA and CB respectively, which divide these sides in the ratio 2 : 1. Prove that

(i) 9 AQ^{2} = 9 AC^{2} + 4 BC^{2}

(ii) 9 BP^{2} = 9 BC^{2} + 4 AC^{2}

(iii) 9 (AQ^{2} + BP^{2}) = 13 AB^{2}

**Sol. **It is given that P divides CA in the ratio 2 : 1. Therefore,

\(CP=\frac { 2 }{ 3 }AC\) ….(i)

Also, Q divides CB in the ratio 2 : 1.

∴ \(QC=\frac { 2 }{ 3 }BC\) ….(ii)

**(i)** Applying pythagoras theorem in right-angled triangle ACQ, we have

AQ^{2} = QC^{2} + AC^{2}

⇒ AQ^{2} = \(\frac { 4 }{ 9 }\) BC^{2} + AC^{2 } [Using (ii)]

⇒ 9 AQ^{2} = 4 BC^{2} + 9 AC^{2 } ….(iii)

**(ii) **Applying pythagoras theorem in right triangle BCP, we have

BP^{2} = BC^{2} + CP^{2}

⇒ BP^{2} = BC^{2} + AC^{2} [Using (i)]

⇒ 9 BP^{2} = 9 BC^{2} + 4 AC^{2} ….(iv)

**(iii) **Adding (iii) and (iv), we get

9 (AQ^{2} + BP^{2}) = 13 (BC^{2} + AC^{2})

⇒ 9 (AQ^{2} + BP^{2}) = 13 AB^{2} [∵ BC^{2} = AC^{2} + AB^{2}]

**Example 11:** In a ∆ABC, AD ⊥ BC and AD^{2} = BC × CD. Prove ∆ABC is a right triangle.

**Sol. **

In right triangles ADB and ADC, we have

AB^{2} = AD^{2} + BD^{2} ….(i)

and, AC^{2} = AD^{2} + DC^{2} ….(ii)

Adding (i) and (ii), we get

AB^{2} + AC^{2} = 2 AD^{2} × BD^{2} + DC^{2}

⇒ AB^{2} + AC^{2} = 2BD × CD + BD^{2} + DC^{2} [∵ AD^{2} = BD × CD (Given)]

⇒ AB^{2} + AC^{2} = (BD + CD)^{2} = BC^{2}

Thus, in ∆ABC, we have

AB^{2} = AC^{2} + BC^{2}

Hence, ∆ABC, is a right triangle right-angled at A.

**Example 12:** The perpendicular AD on the base BC of a ∆ABC intersects BC at D so that DB = 3 CD. Prove that 2AB^{2} = 2AC^{2} + BC^{2}.

**Sol. **We have,

DB = 3CD

BC = BD + DC

The perpendicular AD on the base BC of a ∆ABC intersects BC at D so that DB = 3 CD. Prove that

2AC2 + BC2.

We have,

DB = 3CD

∴ BC = BD + DC

⇒ BC = 3 CD + CD

⇒ BD = 4 CD ⇒ CD = \(\frac { 1 }{ 4 }\) BC

∴ CD = \(\frac { 1 }{ 4 }\) BC and BD = 3CD = \(\frac { 1 }{ 4 }\) BC ….(i)

Since ∆ABD is a right triangle right-angled at D.

∴ AB^{2} = AD^{2} + BD^{2} ….(ii)

Similarly, ∆ACD is a right triangle right angled at D.

∴ AC^{2} = AD^{2} + CD^{2} ….(iii)

Subtracting equation (iii) from equation (ii) we get

AB^{2} – AC^{2} = BD^{2} – CD^{2}

⇒ AB^{2} – AC^{2} = \({{\left( \frac{3}{4}BC \right)}^{2}}-{{\left( \frac{1}{4}BC \right)}^{2}}\) \(\left[ From\ (i)\ CD=\frac{1}{4}BC,\ BD=\frac{3}{4}BC \right]\)

⇒ AB^{2} – AC^{2} = \(\frac { 9 }{ 16 }\) BC^{2} – \(\frac { 1 }{ 16 }\) BC^{2}

⇒ AB^{2} – AC^{2} = \(\frac { 1 }{ 2 }\) BC^{2}

⇒ 2(AB^{2} – AC^{2}) = BC^{2}

⇒ 2AB^{2} = 2AC^{2} + BC^{2}.

**Example 13:** ABC is a right triangle right-angled at C. Let BC = a, CA = b, AB = c and let p be the length of perpendicular from C on AB, prove that

(i) cp = ab

(ii) \(\frac{1}{{{p}^{2}}}=\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}\)

**Sol. (i) **Let CD ⊥ AB. Then, CD = p.

∴ Area of ∆ABC = \(\frac { 1 }{ 2 }\) (Base × Height)

⇒ Area of ∆ABC = \(\frac { 1 }{ 2 }\) (AB × CD) = \(\frac { 1 }{ 2 }\) cp

Also,

Area of ∆ABC = \(\frac { 1 }{ 2 }\) (BC × AC) = \(\frac { 1 }{ 2 }\) ab

∴ \(\frac { 1 }{ 2 }\) cp = \(\frac { 1 }{ 2 }\) ab

⇒ cp = ab

**(ii) **Since ∆ABC is right triangle right-angled at C.

∴ AB^{2} = BC^{2} + AC^{2}

⇒ c^{2} = a^{2} + b^{2}

\(\Rightarrow {{\left( \frac{ab}{p} \right)}^{2}}={{a}^{2}}+{{b}^{2}} \) \(\left[ \because \ cp\,=\,ab\ \ \therefore c=\frac{ab}{p} \right]\)

\(\Rightarrow \frac{{{a}^{2}}{{b}^{2}}}{{{p}^{2}}}={{a}^{2}}+{{b}^{2}} \)

\(\Rightarrow \frac{1}{{{p}^{2}}}=\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}{{b}^{2}}}\Rightarrow \frac{1}{{{p}^{2}}}=\frac{1}{{{b}^{2}}}+\frac{1}{{{a}^{2}}} \)

\(\Rightarrow \frac{1}{{{p}^{2}}}=\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}} \)