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abcd is a rectangle in which diagonal bd bisect |
| Answer» Given: A rectangle ABCD in which diagonal BD bisects {tex}\\angle B{/tex}.\xa0To prove: ABCD is a square.Proof: DC||AB [{tex}\\because{/tex} Opposite sides of a rectangle are parallel]{tex}\\Rightarrow \\;\\angle 4 = \\angle 1{/tex}\xa0..(1) [Alternate interior angles]Similarly,\xa0{tex}\\angle 3 = \\angle 2{/tex} …(2) [Alternate interior angles]And {tex}\\angle 1 = \\angle 2{/tex}…(3) [Given]From equation (1), (2) and (3), we get{tex}\\angle 3 = \\angle 4{/tex}In\xa0{tex}\\Delta BDA{/tex} and\xa0{tex}\\Delta BDC{/tex}, we have{tex}\\angle 1 = \\angle 2{/tex}\xa0[Given]BD = BD [Common side]{tex}\\angle 3 = \\angle 4{/tex}\xa0[proved above]So, By ASA criterion of congruence, we have{tex}\\Delta BDA \\cong \\Delta BDC{/tex}\xa0{tex}\\therefore{/tex} AB = BC [CPCT]So, ABCD is a square.Hence, proved. | |