Question

A is the midpoint of side QR of parallelogram PQRS such that AP bisects angle P.
Prove that PS =2RS

Answer 1

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

Answer 2

Given :- A is the midpoint of side QR of parallelogram PQRS such that AP bisects angle P.

To Prove :- PS = 2•RS .

Proof :-

given that, A is the mid point of side QR .

So,

→ QA = AR -------- Equation (1)

also, given that, AP bisects ∠P .

So,

→ ∠QPA = ∠SPA ----------- Equation (2)

Now, since opposite sides of a parallelogram are parallel to each other .

→ PS is parallel QR and PA is transversal line .

So,

→ ∠PAQ = ∠SPA {Alternate interior angles.} -------- Equation (3)

then, from Equation (2) and Equation (3) we get,

→ ∠QPA = ∠ PAQ -------- Equation (4)

now, in ∆PQA we have,

→ ∠QPA = ∠ PAQ { from Equation (4) }

So,

→ PQ = QA { In a ∆ sides opposite to equal angles are equal in measure .} --------- Equation (5)

now, From (1) and (5) we get,

→ QA = AR = PQ -------- Equation (6)

also,

→ QR = QA + AR

using Equation (6),

→ QR = PQ + PQ

→ QR = 2•PQ ----------- Equation (7)

since opposite sides of a parallelogram are equal in measure .

→ QR = PS

→ PQ = SR

finally putting these values in Equation (7) we get ,

→ PS = 2•SR

→ PS = 2•RS (Proved)

Hence, we can conclude that, PS = 2•RS .

Learn more :-

In the figure along side, BP and CP are the angular bisectors of the exterior angles BCD and CBE of triangle ABC. Prove ∠BOC = 90° - (1/2)∠A .

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