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In Fig. 5.10, if $ \mathrm{AC}=\mathrm{BD} $, then prove that $ \mathrm{AB}=\mathrm{CD} $.
"
Given:
$AC=BD$.
To do:
We have to prove that $AB=CD$.
Solution:
Given,
$AC=BD$
According to the Figure,
We get,
$AC=AB+BC$
$BD=BC+CD$
This implies,
$AB+BC=BC+CD$....(i) (Since, $AC=BD$
According to Euclid's Axiom,
When equals are subtracted from equals, reminders are also equal.
Therefore,
Let us subtract $BC$ from both sides of (i)
We get,
$AB+BC-BC=BC+CD-BC$
This implies,
$AB=CD$
Hence proved.
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