ABCD is a square. A is joined to a point P on BC and D is joined to a point Q on AB. If AP = DQ; prove that AP and DQ are perpendicular to each other.

Let AP and DQ intersect at point O.

In △ DAQ and △ ABP,

⇒ ∠DAQ = ∠ABP (Interior angle of square equal to 90°)

⇒ DQ = AP (Given)

⇒ AD = AB (Each side of square equal in length)

∴ △ DAQ ≅ △ ABP (By R.H.S. congruence rule)

We know that,

Corresponding parts of congruent triangle are equal.

∴ ∠3 = ∠1 ……..(1)

From figure,

⇒ ∠1 + ∠4 = 90°

Substituting value of ∠1 from equation (1) in above equation, we get :

⇒ ∠3 + ∠4 = 90°

In triangle AOD,

By angle sum property of triangle,

⇒ ∠ODA + ∠OAD + ∠AOD = 180°

⇒ ∠3 + ∠4 + ∠AOD = 180°

⇒ 90° + ∠AOD = 180°

⇒ ∠AOD = 180° - 90° = 90°.

∴ AP ⊥ DQ.

Hence, proved that AP and DQ are perpendicular to each other.