solve this problems that can make you genius
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hey mate !!❤✌
here's ur answer ✍✍✍
➡️ DRAW DN PERPENDICULAR TO AC
AND BM PERPENDICULAR TO AC
NOW IN ∆ DON AND ∆BOM
ANGLE DNO = ANGLE BMO { EACH 90°}
ANGLE DON = ANGLE BOM { VERTICALLY OPPOSITE ANGLES }
OD=OB { GIVEN }
➡️ BY AAS CONGRUENCY RULE
∆ DON ≈ ∆ BOM --------(2)
DN = BM ( CPCT )
NOW IN ∆ DNC & ∆ BMA
ANGLE DNC = ANGLE BMA { EACH 90° }
CD=AB { GIVEN }
DN=BM { PROVED ABOVE }
➡️ BY RHS CONGRUENCY RULE
∆ DNC ≈ ∆ BMA -------(1)
✓ ADDING (1)&(2)
=> ∆ DOC ≈ ∆ AOB
AR(DOC)=AR(AOB) { CONGRUENT FIG HAVE EQUAL AREA }
________________________________
AR (∆ DOC) = AR (∆ AOB)
ADDING COB ON BOTH SIDES
=> AR (DOC + COB) = AR ( AOB+ COB)
=> AR (DCB) = AR (ACB)
_______________________________
AR (DCB) = AR (ACB)
IF TWO ∆' ARE ON SAME BASE AND ARE EQUAL IN AREA THEN THEY LIE BETWEEN SAME //s
=> DA // CB
IN QUADRILATERAL ABCD AB=CD AND DA//CB
SO, ABCD IS A //GM ( ONE PAIR OF OPP SIDE IS // AND OTHER IS EQUAL )
______________________________
⭐ Hope it helps ⭐
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here's ur answer ✍✍✍
➡️ DRAW DN PERPENDICULAR TO AC
AND BM PERPENDICULAR TO AC
NOW IN ∆ DON AND ∆BOM
ANGLE DNO = ANGLE BMO { EACH 90°}
ANGLE DON = ANGLE BOM { VERTICALLY OPPOSITE ANGLES }
OD=OB { GIVEN }
➡️ BY AAS CONGRUENCY RULE
∆ DON ≈ ∆ BOM --------(2)
DN = BM ( CPCT )
NOW IN ∆ DNC & ∆ BMA
ANGLE DNC = ANGLE BMA { EACH 90° }
CD=AB { GIVEN }
DN=BM { PROVED ABOVE }
➡️ BY RHS CONGRUENCY RULE
∆ DNC ≈ ∆ BMA -------(1)
✓ ADDING (1)&(2)
=> ∆ DOC ≈ ∆ AOB
AR(DOC)=AR(AOB) { CONGRUENT FIG HAVE EQUAL AREA }
________________________________
AR (∆ DOC) = AR (∆ AOB)
ADDING COB ON BOTH SIDES
=> AR (DOC + COB) = AR ( AOB+ COB)
=> AR (DCB) = AR (ACB)
_______________________________
AR (DCB) = AR (ACB)
IF TWO ∆' ARE ON SAME BASE AND ARE EQUAL IN AREA THEN THEY LIE BETWEEN SAME //s
=> DA // CB
IN QUADRILATERAL ABCD AB=CD AND DA//CB
SO, ABCD IS A //GM ( ONE PAIR OF OPP SIDE IS // AND OTHER IS EQUAL )
______________________________
⭐ Hope it helps ⭐
✔ Mark as brainliest ✔
⚛ Follow me ⚛
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