maths

In $△AMC$ and $△BMD$

$AM=BM$ ....Since $M$ is the midpoint of the hypotenuse $AB$

$CM=DM$ .... Given

$∠AMC=∠BMD$ ..... Vertically opposite angles

$∴AMC≅BMD$ .... SAS Rule

Option B:

$∠ACM=∠BDM$ .... cpct

But these are the alternate angles and they are equal.

$AC∣∣BD$

Now, $AC∣∣BD$ and a transversal $BC$ intersects, then

$∠DBC+∠ACB=180_{o}$

The sum of consecutive interior angles on the same sides of the tranversal is $180_{o}$.

$⇒DBC+90_{o}=180_{o}$

$⇒DBC=90_{o}$

$∴∠DBC$ is a right angle.

Option C:

In $△DBC$ and $△ACB$

$∠DBC=ACB=90_{o}$

$BC=CB$ .... Common side

$∴△AMC≅△BMD$ ... Proved

$AC=BD$ .... cpct

$△DBC≅△ACB$ ..... SAS Rule

Option D:

$△DBC≅△ACB$ .... Proved

$DC=AB$ .... cpct

$⇒2CM=AB$ .... $DM=CM=21 DC$

$⇒CM=21 AB$

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