Question

In ؈m ABCD, T∈ BC and AT intersects BD in M and DC in O. Prove that AM² = MT. MO.

Answer 1

in Parallelogram ABCD , T∈ BC and AT intersect BD in M and DC in O

To prove :- AM² = MT.MO

in ∆AMB and ∆OMD
∠AMB = ∠OMD { vertically opposite angles }
∠MBA = ∠MDO { alternate angles }
from A - A similarity rule,
∆AMB ~ ∆OMD
so, AM/OM = MB/MD --------(1)

similarly, in ∆AMD and ∆TMB
∠AMD = ∠TMB { vertically opposite angles }
∠DAM = ∠MTB { alternate angles}
from A - A similarity rule,
∆AMD ~ ∆TMB
so, AM/TM = MD/MB -------(2)

multiplying equations (1) and (2),
AM/OM × AM/TM = MB/MD × MD/MB
AM²/OM.TM = 1
AM² = OM.TM
therefore , AM² = OM.MT