In the diagram, `AD` is parallel to `BC`, `AC` bisects `/_BAD` and `BD` bisects `/_ABC`. The lines `AC` and `BD` intersect at `P`.
- Prove that `/_BAC = /_BCA`. (1 mark)
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- Prove that `Delta ABP ≡ Delta CBP`. (2 marks)
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Show Worked Solution
| i. |  |
`text(Prove)\ /_BAC = /_BCA`
| `/_BCA` | `= /_CAD\ \ \ text{(alternate angles,}\ BC\ text(||)\ AD text{)}` |
| `/_CAD` | `= /_BAC\ \ \ text{(}AC\ text(bisects)\ /_BAD text{)}` |
| `:. /_BAC` | `= /_BCA\ …\ text(as required)` |
ii. `text(Prove)\ Delta ABP ≡ Delta CBP`
| `/_BAC` | `= /_BCA\ \ \ text{(from part (i))}` |
| `/_ABP` | `= /_CBP\ text{(}BD\ text(bisects)\ /_ABC text{)}` |
| `BP\ text(is common)` | |
`:. Delta ABP ≡ Delta CBP\ \ text{(AAS)}`