Question

$ABCD$ is a quadrilateral in which $\overline{AB}=\overline{AD}$ and $\overline{BC}=\overline{DC}$. Diagonals $\overline{AC}$ and $\overline{BD}$ intersect each other at O.

• A. $\Delta ABC\cong \Delta ADC$
• B. $\Delta AOB\cong \Delta AOD$
• C. $\overline{AC}\perp \overline{BD}$
• D. $\overline{AC}$ bisects $\overline{BD}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $\Delta AOB\cong \Delta AOD$
C $\overline{AC}\perp \overline{BD}$
D $\overline{AC}$ bisects $\overline{BD}$

AB$=$AD(Given)

BC$=$DC(Given)

AC$=$AC(common)

Therefore triangle ABC congruent ACD[SSS]

$\therefore \angle BOA=\angle AOD={90}^o$

$\therefore AB=AD$(Given)

$\therefore AO=AO$(common)

$\angle BOA=\angle AOD={90}^o$

$\therefore \Delta AOB\cong \Delta AOD$

$AC\perp BD$($\because$ABCD is a quadrilateral)

AC bisects BD.

(one diagonal of a quadrilateral bisect the other)

Hence, correct option is B, C, D.