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$βP+βQ+βR=180_{β}80_{β}+40_{β}+βR=180_{β}ββR=180_{β}β120_{β}=60_{β}$

Now $β³PQRββ³ABC$

As corresponding parts of congruent triangle are equal

$β΄AB=PQ=5cm$ and $βC=60_{β}$

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Quadrilateral ABCD is a square. X is the mid-point of AB and Y is the mid-point of BC.Β Hence, $βDXA=βAYB$

**If the above statement is true then mention answer as 1, else mention 0 if false**

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BMDN is a __________.

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(i) AXCY is a parallelogramΒ

(ii) AX=CY,AY=CX

(iii) $β³AYBββ³CXD$

(iv) $β³AXDββ³CYB$

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$β³AMCββ³BMD$

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