Question

In the figure, $AD$ and $BC$ are perpendicular to $AB$, and $AD=BC$. Then , by $SAS$ congruence postulate, $\Delta ABC\cong$

• A. $\Delta ABD$
• B. $\Delta BDA$
• C. $\Delta BAD$
• D. $\Delta ADB$

Answer 1

The correct option is C $\Delta BAD$

$AD\perp AB$ and $BC\perp AB$ [ Given ]

$\therefore$ $\angle BAD=\angle ABC={90}^o$ ---- ( 1 )

In $△ABC$ and $△BAD$

$BC=AD$ [ Given ]

$\angle ABC=\angle BAD={90}^0$ [ From ( 1 ) ]

$AB=BA$ [ Common side ]

$\therefore$ $△ABC\cong △BAD$