solve this q and give me solution it is very important
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i). In ∆DCN and ∆BAP
angle DNC=angle BPA=90°. [A.T.Q.]
DC=AB [oppos. sides of parall. are =]
angle DCN=angle BAP [ DC || BA & AC is transversal,
therefore, by aternate angle property]
So,. ∆DCN=∆BAP ( BY RHS)
ii). In ∆ADN & ∆CBP
AD=CB (oppos. sides of parllelogram are equal)
angleAND=angleCPB (by linear pair it is 90°)
DN=BP ( according to 1st part ∆DCN=∆BAP
and as we know that congruent parts of congruent triangle are equal)
therefore,. ∆ADN=∆CBP (BY SAS)
SO, AN = CP ( congruent parts of congruent triangle are equal)
Hence proved.
angle DNC=angle BPA=90°. [A.T.Q.]
DC=AB [oppos. sides of parall. are =]
angle DCN=angle BAP [ DC || BA & AC is transversal,
therefore, by aternate angle property]
So,. ∆DCN=∆BAP ( BY RHS)
ii). In ∆ADN & ∆CBP
AD=CB (oppos. sides of parllelogram are equal)
angleAND=angleCPB (by linear pair it is 90°)
DN=BP ( according to 1st part ∆DCN=∆BAP
and as we know that congruent parts of congruent triangle are equal)
therefore,. ∆ADN=∆CBP (BY SAS)
SO, AN = CP ( congruent parts of congruent triangle are equal)
Hence proved.
mfazal:
and veeru if you found the congruent parts of the triangle then by CPCT you can then prove that AN =CP
I can’t
There are two people who already answered this question
Sorry only 1 answered and another is answering
Are you in class 9
silly mistake..it will be rhs...srrry
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hey...hope you get the solution.
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You are also wrong
Just use your common scene man
If u are saying that ASA then side should be between the angles
Hey...I'm not on my board exam.
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