Question

Can anyone tell me what is the last condition of congruency for both the questions.

So far I’ve identified 2 each.

Answer 1

In both cases, you have to work on the angle between the sides you found out to be equal.

Diagram 1:

You can show that the angle PCB is the same as ACR. To prove this, you can think of angles:

`P\hat{C}B=P\hat{C}A + A\hat{C}B`

and

`A\hat{C}R=B\hat{C}R+A\hat{C}B`

So, angle
`A\hat{C}B` is common, and both
`P\hat{C}A` and
`B\hat{C}R` are right angles.

So, angles PCB and ACR are both 90° plus a common term (ACB), and thus they're equal.

Diagram 2:

We will work very similarly here: in this case, you have

`A\hat{B}G = A\hat{B}C-G\hat{B}C`

and

`C\hat{B}E=G\hat{B}E-G\hat{B}C`

Again, both
`A\hat{B}C` and
`G\hat{B}E` are right angles, so we have

`A\hat{B}G = 90-G\hat{B}C`

and

`C\hat{B}E=90-G\hat{B}C`

So, both
`A\hat{B}G` and
`C\hat{B}E` are 90 minus a common term, so they are equal.