Theorems And Proofs

IMPORTANT THEOREMS AND PROOFS Theorem: If a perpendicular is drawn from the vertex of the right angle of a right angled triangle to the hypotenuse, the triangle on each side of the perpendicular are similar to the whole triangle and to each other. Given: In a triangle ABC, right angled at B, BD ⊥ AC

To prove: Triangle ADB Similar to Triangle BDC

Triangle ADB Similar to Triangle ABC Triangle BDC Similar to Triangle ABC Proof: (1) In Triangle ABC , ∠B = 90o this implies that ∠ABD + ∠BC = 90o In Triangle DBC ,sin ce∠D = 90o

Therefore ∠DBC + ∠BCD = 90o So, ∠ABD = ∠BCD In Triangle ADB and Triangle BDC we have

∠ABD = ∠BCD and ∠ABD = ∠BCD = 90o

(AA-similarity)

Therefore Triangle ADB Similar to Triangle BDC

(2) In Triangle ADB and Triangle ABC , ∠ABC = ∠ADB = 90o and ∠A is common

Therefore Triangle ADB Similar to Triangle ABC (AA-similarity) (3) In Triangle BDC and Triangle ABC , ∠BDC = ∠ABC = 90o and ∠C is common

Therefore Triangle BDC Similar to Triangle ABC (AA-similarity) Theorem: The ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides. Given: Two triangles, Triangle ABC Triangle ABC Similar to Triangle DEF

and Triangle DEF ,

such

that

AB BC AC = = , ∠A = ∠D, ∠B = ∠E , ∠C = ∠F DE EF DF To prove:

Area of Triangle ABC AB 2 BC 2 AC 2 = = = Area of Triangle DEF DE 2 EF 2 DF 2

Construction: Through A draw AP ⊥ BC and through D draw DQ ⊥ EF

1 1 Proof: Area of Triangle ABC = × BC × AP (area of a Triangle = × base × height 2 2

1 Area Triangle DEF = × EF × DQ 2 1 × BC × AP Area of Triangle ABC 2 BC × AP = = Now 1 Area of Triangle DEF × EF × DQ EF × DQ 2 Area of Triangle ABC ⎛ BC ⎞ ⎛ AP ⎞ =⎜ ⎟ ⎟×⎜ Area of Triangle DEF ⎝ EF ⎠ ⎝ DQ ⎠

… (1)

In Triangle ABP and Triangle DEQ

∠1 = ∠2 = 90o

(Construction)

∠3 = ∠4

(Given)

Therefore Triangle APB Similar to Triangle DQE or

(AA corollary)

AP PB AB = = DQ QE DE AB AP = DE DQ

⇒

or AB BC = DE EF

BC AP (Given) = EF DQ

… (2)

Using (1) and (2), we get

Area of Triangle ABC ⎛ BC ⎞ ⎛ AP ⎞ =⎜ ⎟ ⎟×⎜ Area of Triangle DEF ⎝ EF ⎠ ⎝ DQ ⎠ Area of Triangle ABC ⎛ BC ⎞ ⎛ BC ⎞ BC 2 =⎜ or ⎟×⎜ ⎟= Area of Triangle DEF ⎝ EF ⎠ ⎝ EF ⎠ EF 2 In the same manner, it can be proved that

… (3)

Area of Triangle ABC AB 2 = Area of Triangle DEF DE 2

… (4)

Area of Triangle ABC AC 2 = Area of Triangle DEF DE 2

… (5)

Combining (3), (4) and (5), we get

Area of Triangle ABC AB 2 AC 2 BC 2 = = = Area of Triangle DEF DE 2 DE 2 EF 2 Theorem: (PYTHOGRAS THEOREM): In a right-angled triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides. Given: Triangle ABC is right angled at A, that is ∠BAC = 90o To prove: BC 2 = CA2 + AB 2 Construction: Through A draw AD ⊥ BC Proof: In Triangle ABC and Triangle DBA , we have

∠ABC = ∠ABD

(Common to both the triangles)

∠BAC = ∠ADB = 90o Therefore Triangle ABC Similar to Triangle DBA (AA-criterion of similarity) We know that in similar triangles, corresponding sides are proportional. Therefore

or

AB BC AC = = DB BA DA

BC AB = BA DB

Cross-multiplying both sides, we get AB × BA = BC × DB

or AB 2 = BC × DB

… (1)

Now in Triangle ABC and Triangle ADC

∠CAB = ∠ADC = 90o ∠C = ∠C

(Common to both the triangles)

Therefore By AA criterion of similarity, we have Triangle ABC Similar to Triangle ADC Therefore

Taking

BC AB AC = = AC AD DC

AC BC and cross multiplying, we have = DC AC

AC 2 = BC × DC Adding (1) and (2), we get AB 2 + AC 2 = ( BC × DB) + ( BC × DC ) AB 2 + AC 2 = BC ( DB + DC )

… (2)

AB 2 + AC 2 = BC × BC

[ Since DB + DC = BC ]

AB 2 + AC 2 = BC 2 Theorem: (Converse of Pythagoras theorem): In triangle, if the square o one side is equal to the sum of the squares on the remaining two sides, then the angle opposite t the first side is a right angle. Given: Triangle ABC , such that AB 2 + AC 2 = BC 2 To prove: ∠ACB = 90o that is Triangle ABC , is a right triangle. Construction: Draw a right triangle DEF such that BC=EF, DF=AC and ∠EFD = 90o

Proof: Since ∠EFD = 90o therefore Triangle DEF is right angled triangle. By Pythagoras theorem, we have

DE 2 = EF 2 + FD 2 EF=BC FD=AC

Therefore DE 2 = BC 2 + AC 2

… (1)

Also AB 2 = BC 2 + AC 2

… (2)

So, DE 2 = AB 2

DE = AB Now in Triangle ABC and Triangle DEF , we have AB=DE (proved above) BC=EF (given) AC=DF (given)

Therefore Triangle ABC ≅ Triangle DEF

(SSS-criterion of congruence)

As corresponding parts of congruent triangles are congruent,

Therefore ∠ACB = ∠EFD But ∠EFD = 90o

(By construction)

Therefore ∠ACB = 90o Hence Triangle ABC is right angled triangle, right angled at C. Theorem: The internal bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle. Given: Triangle ABC in which AD is the internal bisector of ∠A and meets BC in D. To prove:

AB BD = AC DC

Construction: Draw CE parallel to DA , meeting BA produced in E.

Proof: Since AD parallel to CE

Therefore ∠2 = ∠3 ∠1 = ∠4

(Alternate angles) and (Corresponding angles)

But ∠1 = ∠2 Since AD bisects ∠A

Therefore ∠3 = ∠4 So AE=AC Now, in Triangle BCE , DA parallel to CE

Therefore

Hence

BA BD = AE DC

AB BD = AC DC

(By Basic Proportionality Theorem)

( Since AE = AC )

Theorem (Converse of Previous Theorem): If a line through one vertex of a triangle divides the opposite sides, in the ratio of other two sides, then the line bisects the angle at the vertex.

Given: Triangle ABC , in which D is a point on BC such that D divides BC in the ratio AB:AC i.e

BD AB = DC AC

To prove: AD is the bisector of ∠A . Construction: Produce BA to E such that AE=AC Proof: In Triangle ACE , we have AE=AC

(by construction)

∠3 = ∠4

… (1)

It is given that

BD AB = DC AC So,

BD AB = DC AE

( Since AC = AE )

Therefore In Triangle BCE , we have BD BA = DC AE

Therefore By converse of basic proportionality theorem we have DA parallel to CE

∠1 = ∠4

(Corresponding angles)

… (2)

and ∠2 = ∠3

(alternate angles)

… (3)

But ∠3 = ∠4

from (1)

Therefore ∠1 = ∠2

From (2) and (3)

Hence AD is the bisector of ∠A . **************************************