Question

in the fig O is the midpoint of each of the line segments AB and CD. Prove that AC=BD and AC||BD

Answer 1

In triangle AOC and BOD
AOC=BOD(VOA)
AO=BO(given)
CO=DO(given)
●∆AOC~=∆BOD(SAS rule)
●AC=BD(cpct)
●AC|| BD(alternate angles will be equal)
Hence proved


Hope it helps u.....

Answer 2

Given: O is the midpoint of AB as well as CD.

To Prove: AC=BD, AC||BD

Step-by-step explanation:

In the figure, we can clearly differentiate two different triangles, $\triangle AOC$ and$\triangle DOB$,

Also, we are given that O is the midpoint of AB and CD, so we can easily say that:

$AO=OB\\CO=OD$                    ...(i)

Now, in  $\triangle AOC$ and$\triangle DOB$:

$AO=OB$                    [by (i)]

$\angle AOC=\angle DOB$         [Vertically opposite angles]

$CO=OD$                    [by (i)]

So, we can say:

$\triangle AOC \cong \triangle BOD$        [By SAS congruency rule]

Now as corresponding parts of congruent triangles are always equal, so:

$AC=BD$                    ...(a)

$\angle OAC=\angle OBD$        

which would mean that  $AC\parallel BD$ as the equal angles act as alternating interior angles.

Thus, $AC\parallel BD$           ...(b)

Hence proved.