Question

he figure, O is the centre of the
le. Seg AB is the diameter.
gent at A intersects the secant
at C. Tangent HD intersects
side AC at J then prove :
1. In the figure
circle. Seg A
Tangent at A in
seg AJ = seg CJ.​

Answer 1

Solution

(Refer attachment for figure)

Given that AB is the diameter of the circle

→ angle BDA = 90°

(Diameter forms right angle at a point of circumference)

Also, we know that radius is perpendicular to tangent

→ angle ODJ = 90°

→ angle OAJ = 90°

Now, let angle ABC = ∅

Then, by angle sun property in ∆ABC, angle DCA becomes 90° - ∅

In ∆BDO,

OD = OB (radius)

So, angle BDO = ∅

Angle ADB = angle BDO + angle ADO

→ 90° = ∅ + angle ADO

→ angle ADO = 90° - ∅

In ∆ODA,

OD = OA (radius)

→ angle ODA = angle OAD = 90° - ∅

Now, angle ODJ = 90°

So, angle ADJ becomes ∅ (refer figure)

And, angle OAJ = 90°

So angle DAJ becomes ∅

→ angle DAJ = angle ADJ

→ AJ = DJ ...... (1)

Again, since angle ADB = 90° and BC is a line, angle ADC = 90° (linear angle)

But angle ADJ = ∅

and angle ADC = 90°

→ angle JDC = 90° - ∅

And we had calculated above that angle DCA = 90° - ∅

→ angle JDC = angle DCA

→ DJ = CJ.......(2)

Equating (1) and (2), we get

AJ = DJ = CJ

→ AJ = CJ

Hence proved.

Answer 2

Answer:

Hey there

Refer to attachment