2025 NYJC P1 Q3

A triangle \(ABC\) is such that \(AB^2 + BC^2 = AC^2\).
By considering \(\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}\) and using the fact that \(|\mathbf{v}|^2 = \mathbf{v} \cdot \mathbf{v}\) for any vector \(\mathbf{v}\), prove that \(\angle ABC\) is a right-angle.[4]

In a triangle \(ABC\), the point \(D\) divides \(AC\) in the ratio \(\lambda : 1 - \lambda\), where \(0 < \lambda < 1\). Let the position vectors of \(A\), \(B\), \(C\) and \(D\) be denoted by \(\mathbf{a}\), \(\mathbf{b}\), \(\mathbf{c}\) and \(\mathbf{d}\) respectively. Show that the area of triangle \(ABD\) is given by\[k|\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a}|\]where \(k\) is to be determined in terms of \(\lambda\).[4]