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In the given figure, altitudes YZ and XT of ∆WXY intersect at P. Prove that,

(1) ▢WZPT is cyclic.

(2) Points X, Z, T, Y are concyclic.

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#### Solution

(1) It is given that, YZ ⊥ WX and XT ⊥ WY.

∴ ∠WZY = 90º .....(1)

∠WTX = 90º .....(2)

Adding (1) and (2), we get

∠WZY + ∠WTX = 90º + 90º = 180º

Or ∠WZP + ∠WTP = 90º + 90º = 180º

In quadrilateral WZPT,

∠WZP + ∠WTP = 180º

We know, if a pair of opposite angles of a quadrilateral is supplementary, then quadrilateral is cyclic.

Therefore, quadrilateral WZPT is cyclic.

(2) It is given that, YZ ⊥ WX and XT ⊥ WY.

∴ ∠XZY = 90º and ∠XTY = 90º

⇒ ∠XZY = ∠XTY

So, two points X and Y on the line XY subtends equal angles at two distinct points Z and T which lie on the same side of the line XY.

Therefore, the points X, Z, T and Y are concyclic.

#### RELATED QUESTIONS

In the following figure, O is the centre of the circle. ∠ABC is inscribed in arc ABC and ∠ ABC = 65°. Complete the following activity to find the measure of ∠AOC.

∠ABC = `1/2`m ______ (Inscribed angle theorem)

______ × 2 = m(arc AXC)

m(arc AXC) = _______

∠AOC = m(arc AXC) (Definition of measure of an arc)

∠AOC = ______

In the above figure, ∠ABC is inscribed in arc ABC.

If ∠ ABC = 60°. find m ∠AOC.

**Solution:**

∠ABC = `1/2`m (arc AXC) ...... `square`

60° = `1/2` m (arc AXC)

`square` = m (arc AXC)

But m ∠ AOC = m(arc`square`) ......(Property of central angle)

∴ m ∠AOC = `square`