Question

please answer my question​

Answer 1

${\large{\pmb{\sf{\underline{RequirEd \; Solution...}}}}}$

${\red{\bigstar}} \:{\pmb{\sf{\underline{Question...}}}}$

⋆ ABCD is a rectangle. AC is a diagonal. By using SSS congruence rule, show that△ABC ≅△CDA.

${\red{\bigstar}} \:{\pmb{\sf{\underline{Given \: that...}}}}$

⋆ ABCD is a rectangle

⋆ AC is a diagonal

${\red{\bigstar}} \:{\pmb{\sf{\underline{To \: proof...}}}}$

⋆△ABC (≅) is congruent to△CDA

${\red{\bigstar}} \:{\pmb{\sf{\underline{Using \: concept...}}}}$

⋆ Congruence rule

${\red{\bigstar}} \:{\pmb{\sf{\underline{Using \: congruence \: rule...}}}}$

⋆ SSS (side side side) congruence rule

${\red{\bigstar}} \:{\pmb{\sf{\underline{Let \: us \: prove...}}}}$

~ As it's given that ABCD is a rectangle and in the rectangle ABCD, AC is a diagonal as drawn in the figure(attachment).

~ As we are able to see in the attachment that when there is a diagonal drawn then it divide this rectangle into two equal parts that is shown like two triangle's.

⟹ ABCD is rectangle (Given)

⟹ AC is diagonal (To show)

⟹ ABC is triangle (Requied)

⟹ CDA is triangle (Requied)

~ Now let's prove!

⟹ AB = DC

⟹ BC = AD

⟹ AC = AC

Henceforth, proved by SSS congruence rule that △ABC ≅ △CDA!