maths

$△AMC≅△BMD$

In right angled $△ABC$,

$∠C=90_{∘}$,

$M$ is the mid-point of $AB$ i.e, $AM=MB$ & $DM=CM$

In $△AMC$ and $△BMD$,

$AM=BM$ ------ (M is the mid-point)

$∠CMA=∠DMB$ ------ (Vertically opposite angles)

$CM=DM$ ------ (Given)

$∴△AMC≅△BMD$, by SAS Postulate

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(1) Equality of two sides of one triangle with any two sides of the second makes the triangle congruent.

(2) Equality of the hypotenuse and a side of one triangle with the hypotenuse and a side of the second respectively makes the triangle congruent.

(3) Equality of the hypotenuse and an acute angle of one triangle with the hypotenuse and an angle of the second respectively makes the triangle congruent.

Of these statements:

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In the following diagrams, ABCD is a square and APB is an equilateral triangle.

In each case,$ΔAPD≅ΔBPC$.

**State whether the above statement is true or false.**

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