Question

Solve this......


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Answer 1

finally I come fast yeh!!

Given, P is the center of the circle and PA and PQ are radius. 

so, In traingle PAQ,

 PA=PQ=r

Therefore, ∠1=∠2 (equal sides have equal angles opposite to them)

Now as PR║AQ,

∠1=∠3 (corresponding angles)

and ∠2=∠4

But, ∠1=∠2 (as proved earlier)

therefore ∠3=∠4

Now inΔPQR and ΔPBR

PQ=PB=r

∠3=∠4 (as proved earlier)

PR=PR (common)

therefore, ΔPQR≡ΔPBR (by SAS rule)

so ∠PQR=∠PBR (corresponding angles of congurent traingle)

here as QR is a tangent, 

therefore ∠PQR=90°

Hence, ∠PBR=90° (as ∠PBR=∠PQR)

Therefore, BR is a tangent at B.

Answer 2

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step-by-step explanation:

First of all,

draw the circle and construction according to the question given.

Now,

use the properties of parallel lines and angles formed by them.

Then,

Use the basic property of triangles like

angles opposite to equal sides are equal.

also,

now use the congruency criteria yo find the relation between the angles and sides.

the CPCT (corresponding part of congruent triangles) is also used

finally,

we know that any line making 90° at the circle is a tangent

this property is used.

all the steps are in attachment.

kindly refer to it.