Question

1. pablo has drawn parallelogram abcd and its diagonals, ( overline{ac} ) and ( overline{bd} ). using side-side-side, he has proven that ( \triangle bad cong \triangle dcb ). given this, which result is pablo now able to prove?

options:

• ( \triangle cob cong \triangle aod )
• the area of ( \triangle boc ) is equal to the area of ( \triangle aob )
• ( angle cob ) is an acute angle
• ( \triangle boc cong \triangle aob )

(diagram of parallelogram ( abcd ) with diagonals intersecting at ( o ) is included)

Explanation:

In a parallelogram \(ABCD\) with diagonals \(AC\) and \(BD\) intersecting at \(O\), the diagonals of a parallelogram bisect each other, so \(AO = OC\) and \(BO=OD\). Also, \(\triangle BOC\) and \(\triangle AOD\): \(BO = OD\) (diagonals bisect), \(AO=OC\) (diagonals bisect), and \(\angle BOC=\angle AOD\) (vertical angles). By SAS congruence, \(\triangle COB\cong\triangle AOD\) is not directly from the given \(\triangle BAD\cong\triangle DCB\), but for the area of \(\triangle BOC\) and \(\triangle AOB\): Since \(AO = OC\) (diagonals bisect), and both triangles \(\triangle BOC\) and \(\triangle AOB\) share the same height from \(B\) to \(AC\). The area of a triangle is \(\frac{1}{2}\times base\times height\). For \(\triangle BOC\), base \(OC\) and for \(\triangle AOB\), base \(AO\), and \(AO = OC\), same height. So their areas are equal. The option about \(\angle COB\) being acute is not necessarily true (could be obtuse), and \(\triangle BOC\cong\triangle AOB\) is not true (they have different side lengths unless it's a rhombus or something). So the correct result is "The area of \(\triangle BOC\) is equal to the area of \(\triangle AOB\)".

Answer:

The area of \(\triangle BOC\) is equal to the area of \(\triangle AOB\) (the option with this statement)