ABCD is a rhombus E, F, G and H are midpoints of AB, BC, CD and DA respectively. Prove that EFGH is a rectangle.
Answers
Step-by-step explanation:
Given- ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively.
To Prove-PQRS is a rectangle
Construction,
AC and BD are joined.
Proof,
In ΔDRS and ΔBPQ,
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 ΔQCR and ΔSAP,
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.
⇒ QR || BD
also,
P and S are the mid points of AD and AB respectively.
⇒ PS || BD
⇒ QR || PS
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.
Given : ABCD is a rhombus
E, F, G and H are midpoints of AB, BC, CD and DA
To Find : Prove that EFGH is a rectangle.
Solution:
ABCD is a rhombus
AB = BC = CD = AD
and AB || CD and BC || AD
E, F, G and H are midpoints of AB, BC, CD and DA
=> FH || AB || CD and FH = AB = CD
Also GE || BC || DA and GE = BC = DA
=> GE = FH ( as AB = BC = CD = AD )
Now using line joining the mid-point of two sides of a triangle is equal to half the length of the third side and Parallel to 3rd side
GH = EF = AC/2 and GH || EF
and FG = EH = BD/2 and EG || FH
Opposite sides are equal and Parallel
and also diagonal are equal in length
Hence Quadrilateral EFGH is Rectangle
QED
Hence Proved
Learn More:
In the adjoining figure, D, E and F are mid-points of the sides BC, CA ...
brainly.in/question/13857685
In the figure, P and Q are the mid - points of the sides AB and BC of ...
brainly.in/question/13016408
