Question

In the given figure, if AB = BC and AD = DC, then ΔABD is congruent to

• A. ΔBDC
• B. ΔBCD
• C. ΔDBC
• D. ΔDCB

Answer 1

The correct option is B ΔBCD

Given: AB = BC and AD = DC
Let AB = BC = x. …..(i)
In ΔABC, by Pythagoras theorem,
$(AC{)}^2=(AB{)}^2+(BC{)}^2$
$\Rightarrow (AC{)}^2=x^2+x^2=2x^2$
$\Rightarrow AC=\sqrt{2}x$
$\therefore AD=DC=\frac{1}{2}AC=\frac{1}{2}\times \sqrt{2}x=\frac{x}{\sqrt{2}}$ .....(ii)
In an isosceles triangle, the median is perpendicular to the base.
Now, AD = DC, ∠ADB = ∠CDB = 90°.
In ΔBDC, by Pythagoras theorem,
$(BC{)}^2=(BD{)}^2+(DC{)}^2$ (Pythagoras theorem)
$\Rightarrow x^2=(BD{)}^2+{\left(\frac{x}{sqrt2}\right)}^2$ [From (i) and (ii)]
$\Rightarrow (BD{)}^2=\frac{x^2}{2}$
$\Rightarrow BD=\frac{2}{\sqrt{2}}$
∴ BD = AD = DC …..(iii)
Now, in ΔABD and ΔBCD,
AB = BC (Given)
BD = CD [From (iii)]
AD = BD [From (iii)]
∴ ΔABD ≅ ΔBCD (SSS congruence rule)
Hence, the correct answer is option (b).