3 Tutor System
Starting just at 265/hour

# Show that the diagonals of a square are equal and bisect each other at right angles.

Given: A square ABCD whose diagonals AC and BD intersect at O.

To prove: Diagonals are equal and bisect each other at right angles. i.e, AC = BD, OD = OB, OA = OC and $${AC}\perp{BD}$$

Proof: In $$\triangle{ABC}$$ and $$\triangle{BAD}$$, we have,
AB = BA ...(Common side)
$$\angle{ABC}$$ = $$\angle{BAD}$$ = $$90^\circ$$
$$\therefore$$ $$\triangle{ABC}$$ $$\displaystyle \cong$$ $$\triangle{BAD}$$ ...(By SAS rule)
$$\therefore$$ AC = BD ...(By CPCT)

Similarly, in $$\triangle{OAB}$$ and $$\triangle{OCD}$$, we have,
AB = DC ...(given)
$$\angle{OAB}$$ = $$\angle{DCO}$$
($$\because$$ AB || CD and transversal AC intersect)
$$\angle{OBA}$$ = $$\angle{BDC}$$
($$\because$$ AB || CD and transversal BD intersect)
$$\therefore$$ $$\triangle{OAB}$$ $$\displaystyle \cong$$ $$\triangle{OCD}$$
(By SAS rule)
$$\therefore$$ OA = OC and OB = OD
(By CPCT)

Now, in $$\triangle{AOB}$$ and $$\triangle{AOD}$$, we have,
OA = OD ...(proved earlier)
(sides of square)
AO = OA
(Commom side)
$$\therefore$$ $$\triangle{AOB}$$ $$\displaystyle \cong$$ $$\triangle{AOD}$$
(By SSS rule)
$$\therefore$$ $$\angle{AOB}$$ = $$\angle{AOD}$$ ...(By CPCT)

$$\angle{AOB}$$ + $$\angle{AOD}$$ = $$180^\circ$$
(linear pair axiom)
$$\angle{AOB}$$ = $$\angle{AOD}$$ = $$90^\circ$$
$$\therefore$$ $$AO\perp{BD}$$
$$\Rightarrow$$ $$AC\perp{BD}$$.
Also, AC = BD, OA = OC, OB = OD and $$AC\perp{BD}$$

Hence, it is proved that diagonals are equal and bisect each other at right angles.