In adjoining figure, T is the mid point of side QR of a parallelogram PQRS such that angle QPT = angle SPT. Prove that PS = 2RS.
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A is the midpoint of side QR of parallelogram PQRS such that AP bisects angle P.Prove that PS =2RS
From the figure we know that PM is the bisector of ∠P
So, we get:
∠QPM=∠SPM…(1)
We know that PQRS is a parallelogram
From the figure, we know that PQ∥SR and PM is a transversal ∠QPM and ∠PMS are alternate angles
∠QPM=∠PMS…(2)
Consider equation (1) and (2):
∠SPM=∠PMS…(3)
We know that the sides opposite to equal angles are equal
MS=PS=9 cm
∠RMT and ∠PMS are vertically opposite angles
∠RMT=∠PMS…(4)
We know that PS∥QT and PT is the transversal
∠RTM=∠SPM
It can be written as
∠RTM=∠RMT
We know that the sides opposite to equal angles are equal
TR=RM
We get:
RM=SR−MS
By substituting the values
RM=12−9
RM=3 cm
RT=RM=3cm
∴,length of RT is 3 cm
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