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(B) $1.$ In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse. Thus,$BM = \frac{1}{2} AC$ $(1-d)

Vedclass · Accepted Answer

$(1-d), (2-c), (3-b), (4-a)$

Vedclass · Answer

(B) $1.$ In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse. Thus,$BM = \frac{1}{2} AC$ $(1-d)$.$2.$ In a right-angled triangle $\Delta ABC$ with altitude $\overline{AD}$ to the hypotenuse,by Apollonius theorem or properties of similar triangles,the relation $AC^2 = CD \cdot BC$ is derived from $\Delta ACD \sim \Delta BCA$ $(2-c)$.$3.$ In a $30^\circ-60^\circ-90^\circ$ triangle,the side opposite to $30^\circ$ is half the hypotenuse. Here,$BC$ is opposite to $30^\circ$,so $BC = \frac{1}{2} AB$ $(3-b)$.$4.$ Apollonius theorem states that for any triangle $\Delta ABC$ with median $\overline{BD}$,$AB^2 + BC^2 = 2(BD^2 + CD^2)$ $(4-a)$.Therefore,the correct match is $(1-d), (2-c), (3-b), (4-a)$.

Part $I$	Part $II$
$1.$ In $\Delta ABC$,$\angle B$ is a right angle and $\overline{BM}$ is a median.	$a. AB^2 + BC^2 = 2(BD^2 + CD^2)$
$2.$ In $\Delta ABC$,$\angle A$ is a right angle and $\overline{AD}$ is an altitude.	$b. BC = \frac{1}{2} AB$
$3.$ In $\Delta ABC$,$m\angle C = 90^\circ$ and $m\angle A = 30^\circ$.	$c. AC^2 = CD \cdot BC$
$4.$ In $\Delta ABC$,$\overline{BD}$ is a median.	$d. BM = \frac{1}{2} AC$

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