DS = BQ (Halves of the opposite sides of the rhombus)
∠SDR = ∠QBP (Opposite angles of the rhombus)
DR = BP (Halves of the opposite sides of the rhombus)
RS = PQ by CPCT --- (i)
RC = PA (Halves of the opposite sides of the rhombus)
∠RCQ = ∠PAS (Opposite angles of the rhombus)
CQ = AS (Halves of the opposite sides of the rhombus)
RQ = SP by CPCT --- (ii)
R and Q are the mid points of CD and BC respectively.
P and S are the mid points of AD and AB respectively.
also, ∠PQR = 90°
RS = PQ and RQ = SP from (i) and (ii)
∠Q = 90°
Copyright © 2024 EHUB.TIPS team's - All rights reserved.
Answers & Comments
Given:-
To Prove:-
Construction:-
Proof:-
In ∆ DRS and ∆ BPQ
DS = BQ (Halves of the opposite sides of the rhombus)
∠SDR = ∠QBP (Opposite angles of the rhombus)
DR = BP (Halves of the opposite sides of the rhombus)
Thus, ΔDRS ≅ ΔBPQ by SAS congruence condition.
RS = PQ by CPCT --- (i)
In ∆ QCR and ∆ SAP,
RC = PA (Halves of the opposite sides of the rhombus)
∠RCQ = ∠PAS (Opposite angles of the rhombus)
CQ = AS (Halves of the opposite sides of the rhombus)
Thus, ΔQCR ≅ ΔSAP by SAS congruence condition.
RQ = SP by CPCT --- (ii)
Now,
In ΔCDB,
R and Q are the mid points of CD and BC respectively.
also,
P and S are the mid points of AD and AB respectively.
Thus, PQRS is a parallelogram.
also, ∠PQR = 90°
Now,
In PQRS,
RS = PQ and RQ = SP from (i) and (ii)
∠Q = 90°
Thus, PQRS is a rectangle.