Question

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.

Answer 1

The encircled part here is the only error rest everything is correct
Hope it helps you
Mark it brainliest pls

Answer 2

YES, 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.