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Question

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

A
ΔBDC
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B
ΔBCD
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C
ΔDBC
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D
ΔDCB
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Solution

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
(AC)2=x2+x2=2x2
AC=2x
AD=DC=12AC=12×2x=x2 .....(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)
x2=(BD)2+(xsqrt2)2 [From (i) and (ii)]
(BD)2=x22
BD=22
∴ 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).

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