PQR is a triangle right angled at P and M is a point on QR
such that PM ⊥ QR. Show that PM² = QM × MR.
PM2
=
QM
MR
×
Proof :
In ΔPQR,
∠QPR
=
90°
PM ⊥ QR
∴
ΔQMP
~
ΔPMR
[If a perpendicular is drawn from the vertex
of the right angle of a right triangle to the
hypotenuse then triangles on both sides of
the perpendicular are similar to the whole
triangle and to each other]
QM
PM
=
PM
MR
[corresponding sides of
similar triangles]
∴
R
Q
P
M
Ex.6.5 (Q.2)
B
A
D
C
ΔBCA
~ ΔBAD
~ ΔACD
Proof :
= BC × BD
∴
∴
AB2
…(i)
ΔBCA
~ ΔBAD
=
BC
BD
BA
BA
In ΔBAD,
m∠BAD
=
90o
AC
⊥
BD
ABD is a triangle right angled at A and AC ⊥ BD.
Show that :
(i) AB2 = BC × BD
(ii) AC2 = BC × CD
(iii) AD2 = BD × CD
∴
[If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of
the perpendicular a are similar to the whole triangle and to each other]
Ex.6.5 (Q.3)
Δ BCA
~ Δ ACD
[From (i)]
Δ BAD
= BD × CD
∴
∴
AD2
[From (i)]
BD
AD
=
AD
CD
[corresponding sides of similar triangle]
~ Δ ACD
ΔBAD ∼ ΔBCA ∼ ΔACD …(i)
= BC × CD
∴
∴
AC2
BC
AC
=
AC
CD
[corresponding sides of similar triangle]
B
A
D
C
Proof :
ABD is a triangle right angled at A and AC ^ BD.
Show that :
(i) AB2 = BC × BD
(ii) AC2 = BC × CD
(iii) AD2 = BD × CD
Ex.6.5 (Q.3)