Question

$\huge\mathbb{\underline{Question:-}}$

In the figure, if AC=BD , then prove that AB= CD ​

Answer 1

Answer:

$\blue{\mathfrak{\underline{\large{Given}}}:} \\ AC=BD \\ \blue{\mathfrak{\underline{\large{To \: prove}}}:} \\ AB = CD \\ \blue{\mathfrak{\underline{\large{Proof}}}:} \\ \boxed{AC=AB+BC}-(i) \\ \boxed{BD=BC+CD}-(i)\\ AC=BD \: (given) \\ \blue{\mathfrak{\underline{\large{By \: {eq}^{n} \: (i) \: and \: (ii) }}}:} \\ AB+BC=BC+CD \\ AB=CD \\ \rightarrow hence,proved \\ \\ \red{\mathfrak{ \large{\underline{{Hope \: It \: Helps \: You}}}}} \\ \blue{\mathfrak{ \large{\underline{{Mark \: Me \: Brainliest}}}}}$

Answer 2

Answer:

AC=AB+BC    ……(ii) (Point B lies between A and C)

BD=BC+CD    … (iii)   (Point C lies between B and D)

Now,  

substituting (ii) and (iii) in (i), we get

⇒AB+BC=BC+CD

⇒AB+BC–BC=CD

⇒AB=CD

Hence, AB=CD.