In a right triangle ABC- angle C-90-circ- M is the midpoint of hypotenuse AB- C is joined to M and produced to a point D such that DM-CM- Point D is joined to point B- triangle AMCcong triangle BMD by-ASARHSSSSSAS

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Toppr Admin · Accepted Answer

In -x25B3-AMC and -x25B3-BMD-x2220-AMC-x2220-BMD -Vertically opposite angles-AM-MB -M is a mid point-DM-CM - -given-Thus- -x25B3-AMC-x2245-x25B3-BMD -by SAS rule-

In a right triangle $A B C,$ $∠ C = 90^{\circ} .$ $M$ is the midpoint of hypotenuse $A B .$ $C$ is joined to $M$ and produced to a point $D$ such that $D M = C M .$ Point $D$ is joined to point $B .$ $△ A M C ≅ △ B M D$ by:

$A S A$
$R H S$
$S S S$
$S A S$

In $△ A M C$ and $△ B M D,$
$∠ A M C = ∠ B M D$ ....(Vertically opposite angles)
$A M = M B$ .....( $M$ is a mid point)
$D M = C M$ .... (given)
Thus, $△ A M C ≅ △ B M D$ ....(by $S A S$ rule)

In a right triangle ABC, ∠C=90∘. M is the midpoint of hypotenuse AB. C is joined to M and produced to a point D such that DM=CM. Point D is joined to point B. △AMC≅△BMD by:ASARHSSSSSAS

In △AMC and △BMD,∠AMC=∠BMD ....(Vertically opposite angles)AM=MB .....(M is a mid point)DM=CM .... (given)Thus, △AMC≅△BMD ....(by SAS rule)

In a right triangle $A B C,$ $∠ C = 90^{\circ} .$ $M$ is the midpoint of hypotenuse $A B .$ $C$ is joined to $M$ and produced to a point $D$ such that $D M = C M .$ Point $D$ is joined to point $B .$ $△ A M C ≅ △ B M D$ by:

$A S A$
$R H S$
$S S S$
$S A S$

In $△ A M C$ and $△ B M D,$
$∠ A M C = ∠ B M D$ ....(Vertically opposite angles)
$A M = M B$ .....( $M$ is a mid point)
$D M = C M$ .... (given)
Thus, $△ A M C ≅ △ B M D$ ....(by $S A S$ rule)