ABCD is a parallelogram BC is proved to L such that BC = CL . AL intersects CD at M . prove that
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Answer:
Given A parallelogram ABCD in which P is a point on side BC such that DP produced meets AB produced at L.
PROOF (i) In △ALD, we have
LB
=
PD
LP
⇒
AB
BL
=
DP
PL
⇒
DC
BL
=
DP
PL
[because AB=DC]
⇒
PL
DP
=
BL
DC
[Taking reciprocals of both sides] [Hence proved]
(ii) From (i), we have
PL
DP
=
BL
DC
⇒
DP
PL
=
DC
BL
[Taking reciprocals of both sides]
⇒
DP
PL
=
AB
BL
[because DC=AB]
⇒
DP
PL
+1=
AB
BL
+1 [Adding 1 on both sides]
⇒
DP
DP+PL
=
AB
BL+AB
⇒
DP
DL
=
AB
AL
⇒
DP
DL
=
DC
AL
[∵AB=DC] [Hence proved]
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