Here we will solve different types of examples on establishing riders based on Pythagoras’ theorem.

**1.** In the quadrilateral PQRS the diagonals PR and QS intersects
at a right angle. Prove that PQ^{2}+ RS^{2} = PS^{2} + QR^{2}.

**Solution:**

Let the diagonals intersect at O, the angle of intersection being a right angle.

In the right-angle ∆POQ, PQ^{2} = OP^{2} + OQ^{2}.

In the right-angle ∆ROS, RS^{2} = OR^{2} + OS^{2}.

Therefore, PQ^{2} + RS^{2} = OP^{2} + OQ^{2} + OR^{2} + OS^{2} ................. (i)

In the right-angle ∆POS, PS^{2} = OP^{2} + OS^{2}.

In the right-angle ∆QOR, QR^{2} = OQ^{2} + OR^{2}.

Therefore, PS^{2} + QR^{2} = OP^{2} + OS^{2} + OQ^{2} + OR^{2} ................. (ii)

From (i) and (ii), PQ^{2}+ RS^{2} = PS^{2} + QR^{2}. (Proved).

**2.** In ∆XYZ, ∠Z = 90° and ZM ⊥ XY, where M is the foot of the perpendicular. Prove that \(\frac{1}{ZM^{2}}\) = \(\frac{1}{YZ^{2}}\) + \(\frac{1}{XZ^{2}}\).

**Solution:**

In ∆XYZ and ∆ZYM,

∠XZY = ∠ZMY = 90°,

∠XYZ = ∠ZYM (Common Angle)

Therefore, by AA criterion of similarity, ∆XYZ ∼ ∆ZYM.

\(\frac{XY}{YZ}\) = \(\frac{XZ}{ZM}\)

⟹ YZ ∙ XZ = XY ∙ ZM

Therefore, ZM = \(\frac{YZ ∙ XZ}{XY}\)

Therefore, \(\frac{1}{ZM^{2}}\) = \(\frac{XY^{2}}{YZ^{2} ∙ XZ^{2}}\) = \(\frac{XZ^{2} + YZ^{2}}{YZ^{2} ∙ XZ^{2}}\); [By Pythagoras’ theorem)

Therefore, \(\frac{1}{ZM^{2}}\) = \(\frac{1}{YZ^{2}}\) + \(\frac{1}{XZ^{2}}\). (Proved)

**3.** In ∆XYZ, ∠Z is acute and XM ⊥ YZ, M being the foot of the perpendicular. Prove that 2YZ ∙ ZM = YZ^{2} + ZX^{2} - XY^{2}.

**Solution:**

From the right-angled ∆XMY,

XY^{2} = XM^{2} + YM^{2}

= XM^{2} + (YZ -
ZM)^{2}

= XM^{2} + YZ^{2}
+ ZM^{2} - 2YZ ∙ ZM (from algebra)

= YZ^{2} - 2YZ ∙
ZM + (XM^{2} + ZM^{2})

= YZ^{2} - 2YZ ∙
ZM + XZ^{2} (from right-angled ∆XMZ)

Therefore, 2YZ ∙ ZM = YZ^{2} + ZX^{2} – XY^{2}. (Proved)

**4.** Let PQRS be a rectangle. O is a point inside the rectangle.
Prove that OP^{2} + OR^{2} = OQ^{2} + OS^{2}.

**Solution:**

PQRS is a rectangle for which PQ = SR = length and QR = PS = breadth.

Join OP, OQ, OR and OS.

Draw XY through O, parallel to PQ.

As ∠QPS and ∠RSP are right angles, ∆PXO, ∆SXO, ∆RYO and ∆QYO are right-angled triangles.

Therefore, by Pythagoras’ theorem,

OP^{2} = PX^{2} + OX^{2},

OR^{2} = RY^{2} + OY^{2},

OQ^{2} = QY^{2} + OY^{2} and

OS^{2} = SX^{2} + OX^{2}

Therefore, OP^{2} + OR^{2} = PX^{2} + OX^{2} + RY^{2} + OY^{2} ......... (i)

OQ^{2} + OS^{2} = QY^{2} + OY^{2} + SX^{2} + OX^{2} ......... (ii)

But in the rectangle XSRY, SX = RY = breadth

and in the rectangle PXYQ, PX = QY = breadth.

Therefore, from (i) and (ii), OP^{2} + OR^{2} = OQ^{2} + OS^{2}.

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