1. ‡KvY ‡f‡` wÎfyR KZ cÖKvi ?
‡KvY †f‡` wÎfyR wZb cÖK

2. wÎfyR
m~²‡KvYx
wÎfyR
mg‡KvYx
wÎfyR
¯’~j‡KvYx
wÎfyR

3. BC AwZfyR
A mg‡KvY
wP‡Î, ABC
GKwU
mg‡KvYx
wÎfyR

4. mg‡KvYx
wÎfy‡Ri
AwZfy‡Ri Dci
Aw¼Z eM©‡¶‡Îi
†¶Îdj Aci `yB
evûi Dci Aw¼Z
eM©‡¶‡Îi †¶Îd‡ji
mg‡KvYx
wÎfy‡Ri GKwU
we‡kl ag© nj t

5. G‡mv, GKwU we‡kl
c×wZ‡Z Gi mZ¨ZvwU
cix¶v Kwi|

6. We first need to show that the angle between
angle x and angle y is a right angle.
•This angle is 90o since x + y = 90o (angle sum of a
triangle) and angles on a straight line add to 180o 
Take 1 identical copy of this right-angled triangle and arrange like so.
Area of trapezium
= ½ (a + b)(a + b) = ½ (a2 +2ab + b2)
Area of trapezium is also equal to the
areas of the 3 right-angled triangles.
= ½ ab + ½ ab + ½ c2
So
½ (a2 +2ab + b2) = ½ ab + ½ ab + ½ c2
a2 +2ab + b2 = 2ab + c2
a2 + b2 = c2 QED
a
b
c
xo
yo
a
bc
xo
yo
Draw line:The boundary shape is a trapezium
To prove that a2 + b2 = c2President James Garfield’s Proof(1876)

7. me© cÖ_g GB
mZ¨wU cÖgvY
K‡ib GKRb MÖxK
`vk©wbK I
MwYZwe`
wc_v‡Mvivm|
‡mBRb¨ GB
Dccv`¨wUi
bvg nq :
Òwc_v‡Mviv‡mi
Dccv`¨Ó

8. wc_v‡Mviv
m †hfv‡e GB
Dccv`¨wU
cÖgvY K‡ib
Zv †`L

9. cieZx©‡Z GB
weL¨vZ
Dccv`¨wU
cÖgvY K‡ib
Avi GK weL¨vZ
MwYZwe`
BDwK¬W

10. B
A
C
D
E
F
G
P Q
g‡b Kwi,
ABC GKwU mg‡KvYx wÎfyR,
hvi A = mg‡KvY
cÖgvY Ki‡Z n‡e †h,
AwZfyR BC Gi Dci Aw¼Z
eM©‡¶‡Îi †¶Îdj = AB evûi
Dci Aw¼Z eM©‡¶‡Îi †¶Îdj
+ AC evûi Dci Aw¼Z eM©‡¶‡Îi
†¶Îdj|
A_©vr, BC2 = AB2 + AC2

11. B
A
C
D
E
F
G
MP Q
L
A¼b t
ABC wÎfy‡Ri ewnf©v‡M BCQP,
ACFG Ges ABDE
eM©‡¶Î wZbwU A¼b Kwi|
A we›`y w`‡q BP Gi mgvš—ivj
AM †iLvsk AvuwK hv
BC †K L Ges PQ †K M we›`y‡Z
†Q` K‡i|
D, C Ges A, P †hvM Kwi|

12. B
A
C
D
E
F
G
MP Q
L
cÖgvY t
BAC = GKmg‡KvY
Ges BAE = GK mg‡KvY
BAC + BA 2 mg‡KvY
∴ CA Ges AE GKB mij‡iLvq
Aew¯’Z|

13. B
A
C
D
E
F
G
MP Q
LAvevi,
CBP = ABD = GK mg‡KvY
Dfqcv‡k ABC †hvM K‡i
CBΡ + ABC ABD + ABC
ABΡ CBD

14. B
A
C
D
E
F
G
MP Q
L
∆ABP I ∆CBD - G
AB BD, BΡ BC
Aš—f©~³ ABΡ = Aš—f©~³ CBD
∆ABΡ ∆CBD
∆‡¶Î ∆ABΡ Gi †¶Îdj = ∆‡¶Î ∆CBD Gi †¶Îdj

15. ∆ †¶Î ABP Ges AvqZ‡¶Î BPML
GKB f~wg BP Gi Dci Ges GKB
mgvš—ivj hyMj BP I AM Gi g‡a¨
Aew¯’Z|
AvqZ‡¶Î BPML Gi †¶Îdj = 2 (∆†¶Î ABP Gi
†¶Îdj)
B
A
C
D
E
F
G
MP Q
L

16. ∆ †¶Î CBD Ges eM©‡¶Î ABD
GKB f~wg BD Gi Dci Ges GKB
mgvš—ivj hyMj BD I C Gi g‡a¨
Aew¯’Z|
eM©‡¶Î ABD Gi †¶Îdj = 2 (∆†¶Î CBD Gi
†¶Îdj)
B
A
C
D
E
F
G
MP Q
L

17. B
A
C
D
E
F
G
MP Q
L
AvqZ‡¶Î BPML Gi †¶Îdj = eM©‡¶Î ABD Gi †¶Îdj

18. GKBfv‡e, A, Q Ges B, F †hvM K‡i cÖgvY Kiv
hvq †h,
AvqZ‡¶Î CQML Gi †¶Îdj = eM©‡¶Î ACFG Gi
†¶Îdj
B
A
C
D
E
F
G
MP Q
L

19. B
A
C
D
E
F
G
P Q
AvqZ‡¶Î BPMLGi †¶Îdj + AvqZ‡¶Î CQML Gi
†¶Îdj
= eM©‡¶Î ABD Gi †¶Îdj + eM©‡¶Î ACFG Gi
†¶Îdj
M
L
A_©vr, eM©‡¶Î BCQΡ Gi †¶Îdj
= eM©‡¶Î ABD Gi †¶Îdj + eM©‡¶Î ∆ACFG Gi

20. BC Gi Dci Aw¼Z eM©‡¶‡Îi †¶Îdj =
AB evûi Dci Aw¼Z eM©‡¶‡Îi †¶Îdj + AC evûi Dci Aw¼Z
eM©‡¶‡Îi †¶Îdj|
A_©vr, BC2 = AB2 + AC2 cÖgvwYZ
B
A
C
D
F
G
P Q
E