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Congruent Figures

D is the mid ...

Question

D is the mid point of the base BC of a triangle ABC. DM and DN are perpendiculars on AB and AC respectively. If $D M = D N$ , the triangle is

A

Isosceles

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B

Equilateral

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C

Right angled

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D

Scalene

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Solution

The correct option is B Isosceles
Given: D is mid point of BC. $D M ⊥ A B$ and $D N ⊥ A C$ , $D M = D N$
Now, In $△ D M B$ and $△ D N C$ ,
$D M = D N$ (Given)
$∠ D M B = ∠ D N C$ (Each $90^{\circ}$ )
$B D = D C$ (D is mid point of BC)
Thus, $△ D M B ≅ △ D N C$ (SAS rule)
Thus, $∠ B = ∠ C$ (By cpct)
hence, $△ A B C$ is an Isosceles triangle.

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