plz solve this question
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In ∆ABD and ∆ABC
BC=AD
angleDAB= angleCBA
AB=AB. ( common side)
By SAS congruence rule
∆ ABD =~∆ABC
BD=AC (CPCT)
angle ABD = angle BAC. ( CPCT)
BC=AD
angleDAB= angleCBA
AB=AB. ( common side)
By SAS congruence rule
∆ ABD =~∆ABC
BD=AC (CPCT)
angle ABD = angle BAC. ( CPCT)
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Given :- ABCD is a quadrilateral in which AD = BC and angle DAB = angle CBA.
To prove :- (i) ∆ABD ≈ ∆BAC
(ii) BD = AC
(iii) angle ABD = angle BAC
Proof :- (i) In ∆ABD and ∆BAC
AD = BC ( given )
angle DAB = angle CBA ( given )
AB = AB ( common )
Therefore, ∆ABD ≈ ∆BAC ( by SAS criterion )
(ii) Since, ∆ABD ≈ ∆BAC ( proved above)
Therefore, BD = AC ( by c.p.c.t. )
(iii) Since, ∆ABD ≈ ∆BAC ( proved above)
Therefore, angle ABD = angle BAC ( by c.p.c.t. )
Given :- ABCD is a quadrilateral in which AD = BC and angle DAB = angle CBA.
To prove :- (i) ∆ABD ≈ ∆BAC
(ii) BD = AC
(iii) angle ABD = angle BAC
Proof :- (i) In ∆ABD and ∆BAC
AD = BC ( given )
angle DAB = angle CBA ( given )
AB = AB ( common )
Therefore, ∆ABD ≈ ∆BAC ( by SAS criterion )
(ii) Since, ∆ABD ≈ ∆BAC ( proved above)
Therefore, BD = AC ( by c.p.c.t. )
(iii) Since, ∆ABD ≈ ∆BAC ( proved above)
Therefore, angle ABD = angle BAC ( by c.p.c.t. )
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