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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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