Math, asked by Rajdeep437, 2 days ago

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.

Attachments:

Answers

Answered by larshikhakrishnan
0

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

Similar questions