Similar Triangles Class 10
Introduction
We studied in the previous class about the congruence of any two geometric figures or any two triangles in particular. We have learnt that congruent figures are alike in all respects. In our daily life, we see many figures or things which have same shape but not necessarily of the same size. Such figures are said to be similar. Our real life is full of examples of similar figures or things; a map of a country and the country itself, a picture on a film and its projection on a screen, a model of a ship and the ship itself, the shadow of an object and the object itself etc. are all examples of pairs of similar figures. When we talk about two similar geometric figures, it is always possible to stretch or shrink one figure in a given scale to get the other figure without changing its shape. In this module we will learn about similar triangles.
Similar triangles
Two triangles ΔABC and ΔDEF are called similar if
(i) Their corresponding angles are equal
i.e. ∠A = ∠D, ∠B = ∠E and ∠C = ∠F
and (ii) Their corresponding sides are proportional
i.e., \frac{{{\rm{AB}}}}{{{\rm{DE}}}}= \frac{{{\rm{BC}}}}{{{\rm{EF}}}} = \frac{{{\rm{AC}}}}{{{\rm{DF}}}}
Figure
If two triangles ΔABC and ΔDEF are similar then we write ΔABC ~ ΔDEF.
Remark
There is a one to one correspondence between the different parts of two similar triangles i.e. if ΔABC ~ ΔPQR then this implies that,
(i) ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R and
(ii) \frac{{{\rm{AB}}}}{{{\rm{PQ}}}} =\frac{{{\rm{BC}}}}{{{\rm{QR}}}} = \frac{{{\rm{AC}}}}{{{\rm{PR}}}}
Some important theorems
Theorem 1. (Basic Proportionality Theorem or Thales Theorem):
If a line is drawn parallel to one side of a triangle, then the other two sides are divided in the same ratio.
Given : ABC is a triangle in which a line DE parallel to BC intersects AB and AC at D and E respectively [Figure] .
[Figure]
To Prove that \frac{{AD}}{{DB}} = . \frac{{AE}}{{EC}}
Construction: We join BE and CD and draw EM⊥ AB and DN ⊥ AC
Proof: We have
\frac{{ar\left( {\Delta ADE} \right)}}{{ar\left( {\Delta DBE} \right)}} =\frac{{\frac{1}{2}AD \times EM}}{{\frac{1}{2}DB \times EM}}[ ∴ Area of a triangle = \frac{1}{2} base × altitude and EM is common altitude ofΔ ADE and Δ DBE]
= \frac{{AD}}{{DB}} …..(i)
Similarly,
\frac{{ar\left( {\Delta ADE} \right)}}{{ar\left( {\Delta CDE} \right)}} = \frac{{\frac{1}{2}AE \times DN}}{{\frac{1}{2}EC \times DN}}[DN is the common altitude of Δ ADE andΔ CDE]
= \frac{{AE}}{{EC}} …..(ii)
Now, ar (ΔDBE) = ar (ΔCDE) …..(iii)
[∴ Δ DBE andΔ CDE stand on the same base DE and lie between the same parallels DE and BC]∴ From (i), (ii) and (iii), we have
=\frac{{ar\left( {\Delta ADE} \right)}}{{ar\left( {\Delta DBE} \right)}} = \frac{{ar\left( {\Delta ADE} \right)}}{{ar\left( {\Delta CDE} \right)}}
⇒ \frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}
[From (i) and (ii)]Corollary
If in a ΔABC, a line DE is drawn parallel to BC intersecting AB and AC at D and E respectively then,
(i) \frac{{{\rm{AB}}}}{{{\rm{DB}}}} = \frac{{{\rm{AC}}}}{{{\rm{EC}}}}
(ii) \frac{{{\rm{DB}}}}{{{\rm{AB}}}} = \frac{{{\rm{EC}}}}{{{\rm{AC}}}}
(iii) \frac{{{\rm{AB}}}}{{{\rm{AD}}}} = \frac{{{\rm{AC}}}}{{{\rm{AE}}}}
(iv) \frac{{{\rm{AD}}}}{{{\rm{AB}}}} = \frac{{{\rm{AE}}}}{{{\rm{AC}}}}
Theorem 2. (Converse of Basic Proportionality Theorem):
If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.
Given : ABC is a triangle in which a line DE intersects AB and AC at D and E respectively such that
\frac{{AD}}{{DB}} = \frac{{AE}}{{EC}} 
Figure
To prove that DE || BC.
Proof: If possible, let DE be not parallel to BC and let another line DF through D be parallel to BC.
Now, since DF || BC, hence, by the Basic Proportionality Theorem, we have
\frac{{AD}}{{DB}}=\frac{{AF}}{{FC}} …..(i)Also, \frac{{AD}}{{DB}}=\frac{{AE}}{{EC}} …..(ii)
[Given]From (i) and (ii), we have
\frac{{AF}}{{FC}}= \frac{{AE}}{{EC}}⇒ \frac{{AF}}{{FC}} + 1 =\frac{{AE}}{{EC}} + 1
[Adding 1 on both sides]⇒ \frac{{AF + FC}}{{FC}}= \frac{{AE + EC}}{{EC}}
⇒ \frac{{AC}}{{FC}}= \frac{{AC}}{{EC}}
⇒ FC = EC
Which is absurd unless F coincides with E.
Hence, our assumption that DE is not parallel to BC is wrong. Hence, DE || BC.
Theorem 3. The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.
Given : ABC is a triangle and AD is the internal bisector of ∠ BAC meeting BC at D [figure]
∴ ∠BAD = ∠ DAC.
Figure
To prove that \frac{{BD}}{{DC}} = \frac{{AB}}{{AC}}
Construction: We draw CE || DA to cut BA produced at E.
Proof: Since AD || EC and AC cuts them, hence we have
∠DAC = ∠ACE …..(i) [Alternate angles ]
and ∠BAD = ∠AEC …..(ii) [Corresponding angles]
But ∠ BAD = ∠DAC [Given]
∴ ∠ACE = ∠AEC [From (i) and (ii)]
∴ AE = AC …..(iii) [∴ Sides opposite to equal angles are equal]
Now, in Δ BCE, we have
CE || DA [By construction]
∴ \frac{{BD}}{{DC}} = \frac{{AB}}{{AE}} [By Basic Proportionality Theorem]
∴ \frac{{BD}}{{DC}} = \frac{{AB}}{{AC}} [From (iii)]
The converse of the theorem is also true.
Theorem 4. (Converse of Theorem 3)
In a triangle ABC, if D is a point on BC such that \frac{{BD}}{{DC}} =\frac{{AB}}{{AC}} , then AD is the internal bisector of∠ BAC.
OR
If a line-segment through one vertex of a triangle divides the opposite side in the ratio of other two sides, then the line bisects the angle at the vertex internally.
Given : ABC is a triangle and D is a point on BC such that
\frac{{BD}}{{DC}} = \frac{{AB}}{{AC}}
Figure
To prove that AD is the internal bisector of ∠ BAC.
Construction: We produce BA to E such that AE = AC. We join CE.
Proof: In Δ AEC, since AE = AC, hence,
∠AEC = ∠ACE …..(i) [ ∴ Angles opposite to equal sides of a triangle are equal]
Now, \frac{{BD}}{{DC}} = \frac{{AB}}{{AC}} [Given]
⇒ \frac{{BD}}{{DC}} = \frac{{AB}}{{AE}} [∴ AE = AC by construction]
∴ By the converse of Basic Proportionality Theorem, we have
DA || CE
Now, since CA is a transversal, we have
∠BAD = ∠AEC …..(ii) [Corresponding angles]
and ∠DAC = ∠ACE …..(iii) [Alternate angles]
But since ∠AEC = ∠ACE, [From (i)]
hence, ∠BAD = ∠DAC [From (ii) and (iii)
Thus, AD bisects ∠BAC internally.
Illustrations
Ex. 1. In the figure , ∠A = ∠B and AD = BE. Show that AB || DE.
Figure
Sol. In Δ ABC, we have
∠A = ∠B [Given]
⇒ AC = BC …..(i) [ ∴ Sides opposite to equal angles of a triangle are equal]
Also, AD = BE ….(ii) [Given]
⇒ AC –CD = BC – CE
⇒ CD = CE ….(iii)
From (ii) and (iii), we have
\frac{{CD}}{{AD}} = \frac{{CE}}{{BE}}
⇒ AB || DE [By converse of Basic Proportionality Theorem]
Ex. 2. ABC is a triangle and D and E are the points on AB and AC such that DE || BC in figure . If \frac{{AD}}{{DB}} =\frac{2}{3} , AC = 18 cm, find AE.
Figure
Sol.
Let AE = x cm
⇒ EC = AC – AE = (18 – x) cm.
Now, since DE || BC, hence, by Basic Proportionality Theorem, we have
\frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}
⇒ \frac{2}{3} = \frac{x}{{18 - x}}
⇒ 2(18 – x) = 3x
⇒ 36 – 2x = 3x
⇒ 2x + 3x = 36
⇒ 5x = 36
⇒ x = \frac{{36}}{5} = 7.2
AE = 7.2 cm.
Ex. 3. In Figure 12.8, D is a point on AB and E is a point on AC of a triangle ABC such that DE || BC. If AD : DB = 1 : 2, AB = 18cm and AC = 30 cm, find AE, ED, AD and DB.
Figure
Sol.
Let AD = x cm and AE = y cm
⇒ DB = AB – AD = (18 – x) cm
and EC = AC – AE = (30 – y) cm
Now, since DE || BC, hence, by Basic Proportionality Theorem, we have
\frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}⇒ \frac{x}{{18 - x}} = \frac{y}{{30 - y}} …..(i)
Also, \frac{{AD}}{{DB}} = \frac{1}{2} …..(ii) [Given]
From (i) and (ii), we have
\frac{x}{{18 - x}} = \frac{y}{{30 - y}} = \frac{1}{2} …..(iii)From the first and the third ratios of (iii), we have
\frac{x}{{18 - x}} = \frac{1}{2}⇒ 2x = 18 – x
⇒ 2x + x = 18 ⇒ 3x = 18 ⇒ x = \frac{{18}}{3} = 6
From the second and the third ratios of (iii), we have
\frac{y}{{30 - y}} = \frac{1}{2}⇒ 2y = 30 – y
⇒ 2y + y = 30 ⇒ 3y = 30 ⇒ y = \frac{{30}}{3} = 10
Hence, AE = 10 cm, EC = (30 – 10) cm = 20 cm,
AD = 6 cm and DB = (18 – 6) cm = 12 cm.
Ex. 4. In figure , AB + AC = 14 cm, AC – AB = 2 cm and D is a point on BC such that BD + DC = 7 cm and DC – BD = 1 cm.
Prove that AD is the internal bisector of ∠ BAC.
Figure
Solution.
We have
AB + AC = 14 cm …..(i)
and AC – AB = 2 cm …..(ii)
Adding (i) and (ii), we get
2AC = (14 + 2) cm = 16 cm
∴ AC = \frac{{16}}{2}cm = 8 cm …..(A)
Subtracting (ii) from (i), we get
2AB = (14 – 2) cm = 12 cm
∴ AB = \frac{{12}}{2}cm = 6 cm …..(B)
Again, BD + DC = 7 cm …..(iii)
and DC – BD = 1 cm …..(iv)
Adding (iii) and (iv), we get
2DC = (7 + 1) cm = 8 cm
∴ DC = \frac{8}{2}cm = 4 cm …..(C)
Subtracting (iv) from (iii), we get
2BD = (7 – 1) cm = 6 cm
∴ BD = \frac{6}{2}cm = 3 cm …..(D)
Now, \frac{{AB}}{{AC}} = \frac{6}{8} [From (A) and (B)]
= \frac{3}{4}
and = \frac{3}{4} [From (C) and (D)]
Hence, \frac{{AB}}{{AC}}= \frac{{BD}}{{DC}}
Hence, AD is the internal bisector of ∠ BAC.
Ex. 5. In the figure , AD is the internal bisector of ∠ A. If AB = 5 cm, BD = 2 cm and BC = 5.4 cm, find AC.
Figure
Sol. We have DC = BC – BD = (5.4 – 2) cm = 3.4 cm
Now, since AD is the internal bisector of ∠BAC,
we have
\frac{{AB}}{{AC}} = \frac{{BD}}{{DC}}
⇒ \frac{5}{{AC}} = \frac{2}{{3.4}}
⇒ 2AC = 3.4 × 5 = 17
⇒ AC = \frac{{17}}{2} = 8.5
Hence, the required length of AC is 8.5 cm
Ex. 6. Prove that the diagonals of a trapezium divide each other proportionally.
Sol. Given that ABCD is a trapezium with AB || CD and AC and DB are its two diagonals interecting at O.
To prove that\frac{{AO}}{{OC}} = \frac{{BO}}{{OD}}
Construction: From O, we draw OE ||AB and OE ||DC intersecting AD at E [Figure].
Figure
Proof: Δ ADB, EO || AB, by construction.
Hence, by Basic Proportionality Theorem, we have
\frac{{DE}}{{EA}}= \frac{{DO}}{{OB}} …..(i)
Again, in Δ ADC, EO || DC, by construction.
Hence, by Basic Proportionality Theorem, we have
\frac{{AE}}{{ED}} = \frac{{AO}}{{OC}} ⇒ \frac{{EA}}{{DE}} = \frac{{AO}}{{OC}} [ ∴ AE = EA and ED = DE]
⇒ \frac{{DE}}{{EA}} = \frac{{OC}}{{AO}} …..(ii)
From (i) and (ii), we have
\frac{{OC}}{{AO}} = \frac{{DO}}{{OB}} = \frac{{OD}}{{BO}}
⇒ \frac{{AO}}{{OC}} =\frac{{BO}}{{OD}}
Ex. 7. If the diagonals of a quadrilateral divide each other proportionally, prove that the quadrilateral is a trapezium.
Sol. Given : ABCD is a quadrilateral and, AC and BD are its two diagonals intersecting at O such that
\frac{{AO}}{{OC}} =\frac{{BO}}{{OD}}
To prove : ABCD is a trapezium with AB || DC.
Construction: We draw a line-segment OE through O parallel to DC to intersect AD at E [Figure].
Figure
Proof: In Δ ADC, we have
EO || DC
∴ \frac{{AO}}{{OC}} = \frac{{AE}}{{ED}} [By Basic Proportionality Theorem] …..(i)
Also, \frac{{AO}}{{OC}} = \frac{{BO}}{{OD}} [Given] …(ii)
∴ From (i) and (ii), we have
\frac{{AE}}{{ED}} = \frac{{BO}}{{OD}}
⇒ \frac{{EA}}{{DE}} = \frac{{BO}}{{OD}}
⇒ \frac{{DE}}{{EA}} = \frac{{DO}}{{OB}} …..(iii)
Thus, the line-segment EO divides the sides AD and BD of Δ ADB such that
\frac{{DE}}{{EA}} =\frac{{DO}}{{OB}} [From (iii)]
Hence, by converse of Basic Proportionality Theorem,
EO || AB.
Also, EO || DC [By construction]
∴ AB || DC
Hence, ABCD is a trapezium.
Ex. 8. Prove that a line-segment drawn parallel to the parallel sides of a trapezium divides the non-parallel sides propotionally.
Given : ABCD be a trapezium with AB || DC, EF is a line-segment parallel to AB and DC [figure]
Figure
To prove that \frac{{AE}}{{ED}} =\frac{{BF}}{{FC}}
Construction: We join AC intersecting EF at G.
Proof: In D ADC, we have
EG || DC.
∴ By Basic Proportionality Theorem,
\frac{{AE}}{{ED}} = \frac{{AG}}{{GC}} …..(i)
Again, in Δ ACB, we have
GF || AB.
∴ By Basic Proportionality Theorem,
\frac{{CG}}{{GA}} =\frac{{CF}}{{FB}}
⇒ \frac{{GC}}{{AG}}=\frac{{CF}}{{FB}}
⇒ \frac{{AG}}{{GC}} =\frac{{FB}}{{CF}} …..(ii)
From (i) and (ii), we have
\frac{{AE}}{{ED}} =\frac{{FB}}{{CF}} = \frac{{BF}}{{FC}}Ex. 9. If three or more parallel lines are intersected by two transversals, prove that the intercepts, made by them on thetransversals are proportional.
Sol. Given that l, m and n are three lines parallel to each other and these lines are intersected by two transversals p and q at A, C, E and B, D, F respectively forming the intercepts AC, CE, BD and DF [Figure].
Figure
To prove that \frac{{AC}}{{CE}}= \frac{{BD}}{{DF}}
Construction: We draw the line-segment AH parallel to the line q meeting the lines m and n at G and H respectively.
Proof: Since AB || GD [Given]
and AG || BD, [By construction]
hence, the figure ABDG is a || gm.
∴ AG = BD …..(i) [ ∴ Opposite sides of a || gm are equal]
Again, GD || HF [Given]
and GH || DF [By construction]
∴ GDFH is a || gm.
∴ GH = DF …..(ii) [∴ Opposite sides of a || gm are equal]
Now, in Δ AEH, CG || EH [Given]
∴ By Basic Proportionality Theorem, we have
\frac{{AC}}{{CE}}=\frac{{AG}}{{GH}} =\frac{{BD}}{{DF}} [From (i) and (ii)]
Ex.10. In Figure , ABC is a triangle and P, Q and R are points on BC, CA and AB respectively such that PQ || AB and QR || BC. Prove that RP || CA, if R is the mid-point of AB.
Sol. Since PQ || AB, hence, we have
\frac{{CQ}}{{QA}} = \frac{{CP}}{{PB}} …..(i) [By Basic Proportionality Theorem]
Figure
Again, since QR || BC, hence, we have
\frac{{AQ}}{{QC}}=\frac{{AR}}{{RB}} [By Basic Proportionality Theorem]
⇒ \frac{{QA}}{{CQ}}=\frac{{AR}}{{RB}}
⇒ \frac{{CQ}}{{QA}} =\frac{{RB}}{{AR}} …..(ii)
From (i) and (ii), we have
\frac{{RB}}{{AR}}= \frac{{CP}}{{PB}}
⇒ \frac{{RA}}{{BR}} =\frac{{PC}}{{BP}} [ RB = RA, R being the mid-point of AB]
⇒ \frac{{BR}}{{RA}} =\frac{{BP}}{{PC}}
Thus, in Δ ABC, we have
\frac{{BR}}{{RA}} = \frac{{BP}}{{PC}}
Hence, by converse of Basic Proportionality Theorem, we have RP || CA.
Ex. 11. In Figure , X is any point inside a triangle ABC. P, Q and R are points on AX, BX and CX respectively such that PQ || AB and QR || BC. Prove that PR || AC.
Figure
Sol In Δ AXB, PQ || AB [Given]
Hence, by Basic Proportionality Theorem, we have
\frac{{XP}}{{PA}} =\frac{{XQ}}{{QB}} …..(i)
Again, in Δ BXC, QR || BC [Given]
Hence, by Basic Proportionality Theorem, we have
\frac{{XQ}}{{QB}}=\frac{{XR}}{{RC}} …..(ii)
From (i) and (ii), we have
\frac{{XP}}{{PA}} =\frac{{XR}}{{RC}}
Thus, in Δ AXC, we have
\frac{{XP}}{{PA}}=\frac{{XR}}{{RC}} .
∴ By converse of Basic Proportionality Theorem, PR || AC.
Ex. 12. Δ ABC is an isosceles triangle in which AB = AC. D and E are the points on AB and AC such that AD = AE. Prove that the points B, C, E and D lie on a circle i.e., they are concyclic.
In other words, prove that the quadrilateral BCED is a cyclic quadrilateral.
Sol. Since AB = AC and AD = AE, [figure]
Figure
∴ \frac{{AD}}{{AB}} =\frac{{AE}}{{AC}}
⇒ \frac{{AB}}{{AD}} =\frac{{AC}}{{AE}}
⇒ \frac{{AB}}{{AD}} – 1 =\frac{{AC}}{{AE}} – 1
⇒ \frac{{AB-AD}}{{AD}} =\frac{{AC-AE}}{{AE}}
⇒ \frac{{BD}}{{AD}}= \frac{{CE}}{{AE}}
⇒ \frac{{AD}}{{BD}}=\frac{{AE}}{{CE}}
Hence, by the converse of Basic Proportionality Theorem, we get
DE || BC.
∴ ∠BDE + ∠DBC = 180º [ BD is a transversal and ∠ BDE and ∠ DBE are the interior angles on the same side of the transversal]
⇒ ∠BDE + ∠BCE = 180º [ ∠DBC = Ð BCE for AB = AC]
Thus, the sum of opposite angles of the quadrilateral BCED is 180º. Hence, B, C, E, D are concyclic.
Ex.13. O is any point inside a triangle ABC. The bisectors of ∠AOB, ∠BOC and ∠COA meet the sides AB, BC and CA in points D, E and F respectively. Show that AD × BE × CF = DB × EC × FA.
Sol. Since OD is the internal bisector of ∠AOB of Δ AOB, we have
\frac{{AO}}{{BO}} =\frac{{AD}}{{DB}} …..(i)
Again, since OE is the internal bisector of ∠BOC of Δ BOC, we have
\frac{{BO}}{{CO}} = \frac{{BE}}{{EC}} …..(ii)
Figure
Finally, since OF is the internal bisector of ∠ AOC of Δ AOC, we have
\frac{{CO}}{{AO}} = \frac{{CF}}{{FA}} …..(iii)
Multiplying (i), (ii) and (iii), we get
\frac{{AO}}{{BO}}×\frac{{BO}}{{CO}} ×\frac{{CO}}{{AO}} =\frac{{AD}}{{DB}} ×\frac{{BE}}{{EC}} ×\frac{{CF}}{{FA}}
⇒ 1 = \frac{{AD \times BE \times CF}}{{DB \times EC \times FA}}
⇒ AD × BE × CF = DB × EC × FA.
Criteria for similarity of two triangles
AAA similarity criterion
If two triangles are equiangular (i.e., if the corresponding angles of two triangles are equal), then their corresponding sides are proportional and hence the two triangles are similar. i.e. In figures (i) & (ii) if ∠A = ∠D, ∠B = ∠E and ∠C = ∠F.
then, ΔABC ~ ΔDEF.
(i) (ii)
Figure
AA similarity criterion
If two angles of one triangle are equal to the corresponding angles of another triangle, then the two triangles are similar.
SSS similarity criterion
If the corresponding sides of two triangles are proportional, then their corresponding angles are equal and hence the two triangles are similar. i.e. In figures (i) and (ii) if \frac{{{\rm{AB}}}}{{{\rm{DE}}}} = \frac{{{\rm{BC}}}}{{{\rm{EF}}}} = \frac{{{\rm{CA}}}}{{{\rm{DF}}}}
then, ΔABC ~ ΔDEF.
(i) (ii)
Figure
SAS similarity criterion
If one angle of a triangle is equal to one angle of the other and the sides including these angles are proportional, then the two triangles are similar to each other. i.e. If in figures (i) & (ii) two triangles ABC and DEF are such that ∠A = ∠D and \frac{{{\rm{AB}}}}{{{\rm{DE}}}}= \frac{{{\rm{AC}}}}{{{\rm{DF}}}}.
then, ΔABC ~ ΔDEF.
(i) (ii)
Figure
Areas of Two Similar Triangles
Theorem 1. The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
ΔABC and ΔPQR are two triangles such that ΔABC ~ ΔPQR.
Then \frac{{ar\left( {\Delta ABC} \right)}}{{ar\left( {\Delta PQR} \right)}} = \frac{{A{B^2}}}{{P{Q^2}}} = \frac{{B{C^2}}}{{Q{R^2}}} = \frac{{C{A^2}}}{{R{P^2}}}
(i) (ii)
Figure
Theorem 2.
If the areas of two similar triangles are equal, then the triangles are congruent.
ΔABC and ΔPQR are two similar triangles and their areas are equal.
Then ΔABC \cong ΔPQR
(i) (ii)
Figure
Pythagoras theorem
Theorem:1 [Pythagoras’ theorem or Baudhayan theorem]:
In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. ABC is a right-angled triangle where ∠BAC = 90º
Then BC2 = AB2 + AC2
Figure
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 first side is a right angle. ABC is a triangle such that AC2 = AB2 + BC2.
(i) (ii)
Figure
Then ∠ ABC = 90º
Important results derived from Pythagoras theorem
Result 1. In figure, Δ ABC is an obtuse-angled triangle, where ∠ ABC is an obtuse angle. If AD is perpendicular to CB produced, then AC2 = AB2 + BC2 + 2 BC.BD [Obtuse triangle property]
Figure
Result 2. In figure, ∠ ABC of Δ ABC is acute and AD ⊥ BC. Then AC2 = AB2 + BC2 – 2 BC.BD. [Acute triangle property]
Figure
Result 3. If AD is a median of a triangle ABC, then AB2 + AC2 = 2 BD2 + 2 AD2
or
In any triangle, the sum of the squares of any two sides is equal to twice the square of half the third side together with twice the square of the median which bisects the third side.
[This result is known as Apollonius’ theorem]
ABC is a triangle and AD is a median of the triangle so that BD = DC. We assume that AB > AC.
Then AB2 + AC2 = 2 BD2 + 2 AD2.
Figure
Result 4. Three times the sum of the squares of the sides of a triangle is equal to four times the sum of the squares of the medians of the triangle.
ABC is a triangle in which AD, BE and CF are the medians.
Then 3 (AB2 + BC2 + CA2) = 4 (AD2 + BE2 + CF2)
Figure
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