Theorem 8.9 :
STaTemenT : The line SegmenT joining The mid-poinTS of
Two SideS of a Triangle
iS parallel To The Third Side.
A

D

B

E

C
giVen:- a Triangle aBC in whiCh d & e are
The mid-poinTS of SideS aB & BC
reSpeCTiVelY SuCh ThaT ad=Bd,
ae=Ce
To proVe: de=1/2BC,de//BC
ConSTruCTion: exTend de To f SuCh ThaT de=ef....(i),join
A
fC

D

B

E

F

C
Proof: Δ AED & Δ CEF
DE=EF
∠ AED = ∠ CEF
AE=EC
Δ AED ≅ Δ CEF
∠DAE= ∠ ECF
AD=CF

(by eq no i)
(V.O.A)
(given)
(by SAS)
(by CPCT)......(ii)
(by CPCT)......(iii)

But ∠ DAE &∠ ECF are A.I.A
∴AB//CF
∴ED//CF
=> AD=BD ……….(iv)(given)
from eq (iii) & (iv)
BD=CF
Therefore, BD=CF,BD//CF
⇒BDCF is //gm
∴DE//BC (DE is part of DF)
⇒DF=DE + EF
= DF=DE + DE (DE=EF)
= DF=2DE
= 1/2DF=DE
= 1/2BC=DE (opp. Sides of //gm are equal)
=>1/2DE=BC,DE//BC
Hece,Proved
Theorem 8.10
STaTemenT: The line drawn Through The midpoinT of one Side of a Triangle, parallel To
anoTher Side biSecTS The Third Side.
A

D

B

E

C
giVen:- a Triangle abc in which d iS midpoinTS of Side ab reSpecTiVely ,Such ThaT
ad=bd , de//bc
To proVe: ae=ec

conSTrucTion: draw cm//ab To meeT de aT f
A
M

D

B

E

F

C
Proof: Δ AED & Δ CEF
∠ AED = ∠ CEF
(V.O.A)
∠ ADE= ∠ EFC
(A.I.A)
AD=FC
(∴ DB=FC,DB=DA)
Δ AED ≅ Δ CEF
(by ASA)
⇒AE=CE
⇒ Hence,Proved
Midpt Theorem & Converse Of Mid pt heorem
Midpt Theorem & Converse Of Mid pt heorem
Midpt Theorem & Converse Of Mid pt heorem

Midpt Theorem & Converse Of Mid pt heorem

  • 2.
    Theorem 8.9 : STaTemenT: The line SegmenT joining The mid-poinTS of Two SideS of a Triangle iS parallel To The Third Side. A D B E C
  • 3.
    giVen:- a TriangleaBC in whiCh d & e are The mid-poinTS of SideS aB & BC reSpeCTiVelY SuCh ThaT ad=Bd, ae=Ce To proVe: de=1/2BC,de//BC ConSTruCTion: exTend de To f SuCh ThaT de=ef....(i),join A fC D B E F C
  • 4.
    Proof: Δ AED& Δ CEF DE=EF ∠ AED = ∠ CEF AE=EC Δ AED ≅ Δ CEF ∠DAE= ∠ ECF AD=CF (by eq no i) (V.O.A) (given) (by SAS) (by CPCT)......(ii) (by CPCT)......(iii) But ∠ DAE &∠ ECF are A.I.A ∴AB//CF ∴ED//CF => AD=BD ……….(iv)(given) from eq (iii) & (iv) BD=CF Therefore, BD=CF,BD//CF ⇒BDCF is //gm ∴DE//BC (DE is part of DF) ⇒DF=DE + EF = DF=DE + DE (DE=EF) = DF=2DE = 1/2DF=DE = 1/2BC=DE (opp. Sides of //gm are equal) =>1/2DE=BC,DE//BC Hece,Proved
  • 5.
    Theorem 8.10 STaTemenT: Theline drawn Through The midpoinT of one Side of a Triangle, parallel To anoTher Side biSecTS The Third Side. A D B E C
  • 6.
    giVen:- a Triangleabc in which d iS midpoinTS of Side ab reSpecTiVely ,Such ThaT ad=bd , de//bc To proVe: ae=ec conSTrucTion: draw cm//ab To meeT de aT f A M D B E F C
  • 7.
    Proof: Δ AED& Δ CEF ∠ AED = ∠ CEF (V.O.A) ∠ ADE= ∠ EFC (A.I.A) AD=FC (∴ DB=FC,DB=DA) Δ AED ≅ Δ CEF (by ASA) ⇒AE=CE ⇒ Hence,Proved