Challenge if you will Solve I will follow u, thank u and Mark as brainlist... super hard question
Challenge if you will Solve I will follow u, thank u and Mark as brainlist... super hard question
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:- P, Q, R and S ae the mid points of side AB, BC, CD and DA pf rhombus ABCD.
:- ABCD is a rectangle.
:- Join BD.
:-
In ∆ ABD
P is the mid point of AB.
and S is the mid point of AD.
So, PS ll BD .......(1)
And PS = ½ BD ........(2)
Also in ∆ BDC.
R and Q are the mid points of sides CD and BC.
So, RQ ll BD .......(3)
And RQ = ½ BD ........(4)
From (1) and (3)
PS ll RQ
And from (2) and (4)
PS ll RQ
In PQRS
PS ll RQ and PS = RQ.
So, PQRS is a parallelogram.
* What we have to prove is that PQRS is a rectangle.
Now,
DC = BC [As sides of rhombus are equal]
Then, their half's are also equal.
½ DC = ½ BC
RC = CQ [As they are mid points of DC and BC]
Now, In ∆ RQC.
RC = CQ
Angle 1 = Angle 2 [Opposite sides of parallelogram are equal] .......(5)
In ∆SDR and ∆ BQP
DR = BQ [DC = BC; ½ DC = ½ BC]
DS = BP [AD = AB; ½ AD = ½ AB]
SR = PQ [Opposite sides of parallelogram are equal]
∆ SDR ~ ∆ BQP
Angle 3 = Angle 4 ......(6)
By Cpct.
Now, DC is a line.
Then,
Angle 3 + Angle SRQ + Angle 1 = 180° .......(7)
Similarly, BC is also a line.
Angle 2 + Angle RQP + Angle 4 = 180°
Angle 1 + Angle RQP + Angle 3 = 180° .....(8)
[From (5) and (6)]
Now, add (7) and (8)
Angle 1 + Angle RQP + Angle 3 = Angle 3 + Angle SRQ + Angle 1
Angle RQP = Angle SRQ .......(9)
Now,
SR ll PQ and RQ is a transversal.
Angle SRQ + Angle RQP = 180°
Angle SRQ + Angle SRQ = 180° [From (9)]
2 (Angle SRQ) = 180°
Angle SRQ = 90°
So, PQRS is a parallelogram with angle 90°.
So,
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