The diagram shows a right-angled triangle `ABC` with `∠ABC = 90^@`. The point `M` is the midpoint of `AC`, and `Y` is the point where the perpendicular to `AC` at `M` meets `BC`.
Show that `\Delta AYM \equiv \Delta CYM`. (2 marks)
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`text(In)\ ΔAYM\ text(and)\ ΔCYM`
| `∠AMY` | `= ∠CMY = 90^@\ \ \ (MY ⊥ AC)` |
| `AM` | `=CM\ \ \ text{(given)}` |
| `YM\ text(is common)` | |
`:. \Delta AYM \equiv \Delta CYM\ \ text{(SAS)}`