Question

In fig o is tge centre of the circle seg ab is the diameter tangent at a intersects the secant bd at c tangent hd intersects side ac at j then prove : seg aj congruent seg cj

Answer 1

Answer: AJ = CJ

Step-by-step explanation:

In the figure,

We can see the 2 tangents going from J to the points A and D.

It is a property of circles, that Tangents from a point to a circle are of equal lengths.

So, AJ = DJ

We can see ABC is a right angle triangle (Radius is always perpendicular to a tangent)

So, Angle ACB + ABC = 90

Triangle ADB is also a right angle triangle, (cords from diameter to a point forms Right Angle)

So, Angle ABD + DAB = 90

Now, Angle BAD + DAC = BAC = 90

We previously proved Angle DAB = ACD

So, in Triangle ADC

Angle DAC + ACD = 90

Therefore, Angle ADC = 90 (Sum of angles of a Triangle)

Angle JAD = JDA (PROVED EARLIER)

As, Angle JDC + JDA = 90

So, Angle JDC = JCD

This gives us, JD = JC

But, JD = AJ

Therefore, AJ = CJ

Hence proved.