Can I write this as:- in ∆ABC, P and Q are the midpoints of AB and BC respectively,
:. PQ||AC and PQ=1/2AC.(mid point theorem)
in ∆ADC, S and R are the midpoints of AB and CD respectively,
:. SR||AC and SR=1/2AC( mid point theorem)
now,SR=1/2AC=PQ.
:. SR=PQ. --(i)
Similarly, by mid point theorem, we can prove that SP=RQ. --(ii)
From (i) & (ii), we get:-
PQ=SR, SP=RQ.
in ∆ASP and in ∆SDR,
AS=SD (halves of sides are equal)
DR=AP ( halves of sides are equal)
∠A=∠C (each∠=90°)
∆ASP≡∆RCQ( SAS CONGRUENCE RULE )
PS=SR (byCPCT). --(iii)
Similarly we can prove PQ=RQ. --(iv)
From (i),(ii),(iii)&(iv) we obtain:-
PQ=QR=RS=SP.
∴ Quadrilateral PQRS is a rhombus.
if yes or no, give reasons.
:. PQ||AC and PQ=1/2AC.(mid point theorem)
in ∆ADC, S and R are the midpoints of AB and CD respectively,
:. SR||AC and SR=1/2AC( mid point theorem)
now,SR=1/2AC=PQ.
:. SR=PQ. --(i)
Similarly, by mid point theorem, we can prove that SP=RQ. --(ii)
From (i) & (ii), we get:-
PQ=SR, SP=RQ.
in ∆ASP and in ∆SDR,
AS=SD (halves of sides are equal)
DR=AP ( halves of sides are equal)
∠A=∠C (each∠=90°)
∆ASP≡∆RCQ( SAS CONGRUENCE RULE )
PS=SR (byCPCT). --(iii)
Similarly we can prove PQ=RQ. --(iv)
From (i),(ii),(iii)&(iv) we obtain:-
PQ=QR=RS=SP.
∴ Quadrilateral PQRS is a rhombus.
if yes or no, give reasons.
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