Question

Triangles ABC and DCB are such that $ar\left(△ABC\right)=ar\left(△DCB\right)$, where A and D lie on the opposite sides of BC. Which of the following options are correct?

• A. $△AMO\cong △DNO$
• B. BC bisects AD
• C. $AO=\frac{1}{2}AD$
• D. None of the above.

Answer 1

Answer: (A), (B), (C)

The correct options are
A $△AMO\cong △DNO$
B BC bisects AD
C $AO=\frac{1}{2}AD$

Before we proceed, let us draw perpendiculars AM and DN from A and D on BC as shown below.
Given that $ar\left(△ABC\right)=ar\left(△DCB\right)$ where BC is the common side to the triangles.
We know that triangles with same base and equal areas have equal corresponding altitudes.
Using this, we have $AM=DN$
Consider triangles AMO and DNO.
$\angle AMO=\angle DNP={90}^{\circ }$
$\angle AOM=\angle DON\left(\text{Vertically opposite angles}\right))$, and
$AM=DN$
Thus, by AAS congruency criterion, we have $△AMO\cong △DNO$.
$⟹AO=DO$, i.e. BC bisects AD.
$⟹AD=AO+OD=AO+AO=2AO⟹AO=\frac{1}{2}AD$