EX.6.4 (Q.3)
ABC and DBC are two triangles on the same base BC.
If AD intersects BC at O,
ar(ΔDBC)
=
AO
ar(ΔABC)
.
DO
show that :
A
D
C
B
O
O
Draw AL ⊥ BC
Construction :
and DM ⊥ BC.
Proof.
=
ar
1
2
BC
×
AL
×
1
2
BC
×
DM
×
(ΔABC)
ar
(ΔDBC)
L
M
In ΔALO and ΔDMO
.... [each 90°, construction]
∠ALO
=
∠DMO
.... [vertically opposite angles]
∠AOL
=
∠DOM
∴
ΔALO
.... [by AA similarity criterion]
~
ΔDMO
.... (i)
=
AL
DM
(ΔABC)
(ΔDBC)
ar
ar
ar (ΔDBC)
ar (ΔABC)
=
AO
DO
.... from (i) & (ii)
∴
AO
DO
=
AL
DM
.... (ii)
.... [corresponding sides of similar triangles.]
D
AL
DM
(ΔABC)
(ΔDBC)
ar
ar
=
.... (i)
ΔALO
~
ΔDMO
Proof.
EX.6.4 (Q.3)
ABC and DBC are two triangles on the same base BC.
If AD intersects BC at O,
ar(ΔDBC)
=
AO
ar(ΔABC)
.
DO
show that :
A
D
C
B
O
L
M