Question

AB and CD bisect each other at K prove that AC = BD.

Answer 1

Given:

AB and CD bisect each other at K.

To Prove:

AC = BD

Calculation:

We will use congruence between two triangles to prove that AC = BD;

In ∆ACK and ∆BKD:

$\rm{1) \: CK=DK \: \: \: \: \: \: .....(K \: bisects \:CD)}$

$\rm{2) \: AK = BK \: \: \: \: ....(K \: bisects \: AB)}$

$\rm{3) \: \angle AKC = \angle BKD \: \: \: ....(opposite \: angles)}$

So, ∆ACK $\cong$ ∆BKD [S-A-S congruency]

So, we can say:

$\boxed{\sf{\large{AC = BD}}}$

( Congruent Parts of Congruent ∆)

[Hence Proved]

Answer 2

Answer:

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