Question

The image of an object placed at a point A before
a plane mirror LM is seen at the point B by an
observer at D as shown in the figure. Prove that
the image is as far behind the mirror as the object
is in front of the mirror. i.e. TA = TB.
​

Answer 1

Answer:

yes TA=TB

Step-by-step explanation:

let take ∆TAC and ∆TBC

by proving these angle conguant we can prove that TA=TB

Answer 2

Step-by-step explanation:

According to the figure we need to prove that AT=BT

We know that

Angle of incidence = Angle of reflection

So we get

∠ACN=∠DCN..(1)

We know that AB∥CN and AC is the transversal

From the figure we know that ∠TAC and ∠ACN are alternate angles

∠TAC=∠CAN(2)

We know that AB∥CN and BD is the transversal

From the figure we know that ∠TBC and ∠DCN are corresponding angles

∠TBC=∠DCN..(3)

By considering the equation (1),(2) and (3)

We get

∠TAC=∠TBC(4)

Now in △ACT and △BCT

∠ATC=∠BTC=90

∘

CT is common i.e. CT=CT

By AAS congruence criterion

△ACT≅△BCT

AT=BT(c.p.c.t)

Therefore, it is proved that the image is as far behind the mirror as the object is in front of the mirror.