Question

in a triangle pqr is right angled at Q qx perpendicular on PR XZ perpendicular on RQ and XZ perpendicular on pq are drawn prove that xz^2=pz×zq

Answer 1

Hi there,  

I think your question has some incorrect information. Considering the correct question to be as follows:

In a triangle pqr is right angled at Q, QX perpendicular on PR, XZ perpendicular on PQ and XY perpendicular on QR are drawn prove that XZ²=PZ×ZQ.

Answer:

Step 1:

In rt. angled ∆PQR,  

Angle Q = 90° …. [given]

Also, in quadrilateral XYQZ, since we are given that,

XZ ⊥ PR and XY⊥ QR

∴ ∠XZQ = ∠XYQ = ∠Q = 90° ….. (i)

Also, ∠ZXY = 90° ….. [since from (i), in a quadrilateral, if 3 of its angles are equal to 90° then the forth angle should also be equal to 90°]

∴ quadrilateral XYQZ is a rectangle [since a rectangle is a quadrilateral with four rt. angles]

Step 2:

Now,  

In ∆ QZX, ∠ZXQ + ∠ZQX = 90° …. (ii)

And,  

In ∆ PZX, ∠ZPX + ∠ZXP = 90° …. (iii)

And,

XQ ⊥ PR, ∠ZXQ + ∠ZXP = 90° …. (iv)

From (ii) & (iv), we get

∠ZQX = ∠ZXP …. (v)

And,

From (iii) & (iv), we get

∠ZPX = ∠ZXQ …. (vi)  

Step 3:

In ∆ QZX and ∆ PZX, we have

XZ [common side]

Angle ZQX = angle ZXP [from (v)]

Angle ZPX = angle ZXQ [from (vi)]

∴ By AAS criterion  

∆ QZX ~ ∆ PZX

∴ PZ/XZ = XZ/ZQ [∵ corresponding sides of two similar triangles are proportional]

∴ XZ² = PZ * ZQ

Hence proved

Hope it helped!!!!