In right triangle ABC, right angledat C. M is the mid-point of hypotenuse AB. C is joined to $ \mathrm{M} $ and produced to a point $ \mathrm{D} $ such that $ \mathrm{DM}=\mathrm{CM} $. Point $ \mathrm{D} $ is joined to point $ \mathrm{B} $. Show that:
(i) $ \triangle A M C \cong \triangle B M D $
(ii) $ \angle \mathrm{DBC} $ is a right angle
(iii) $ \triangle \mathrm{DBC} \cong \triangle \mathrm{ACB} $
(iv) $CM=\frac{1}{2}AB$
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Given:
In right triangle ABC, right angledat C. M is the mid-point of hypotenuse AB. C is joined to \( \mathrm{M} \) and produced to a point \( \mathrm{D} \) such that \( \mathrm{DM}=\mathrm{CM} \). Point \( \mathrm{D} \) is joined to point \( \mathrm{B} \).
To do:
We have to show that
(i) \( \triangle A M C \cong \triangle B M D \)
(ii) \( \angle \mathrm{DBC} \) is a right angle
(iii) \( \triangle \mathrm{DBC} \cong \triangle \mathrm{ACB} \)
(iv) $CM=\frac{1}{2}AB$
Solution:
$\angle ACB=90^o$
$AM=BM$
$DM=CM$
(i) In $\triangle AMC$ and $\triangle DMB$,
$AM=BM$
$CM=DM$
$\angle AMC=\angle BMD$ (Vertically opposite angles)
Therefore,
$\triangle AMC \cong\ \triangle DMB$ (By SAS congruence rule)
(ii) $\triangle AMC \cong\ \triangle DMB$
This implies,
$\angle ACM=\angle BDM$ (CPCT)
$\angle ACM$ and $\angle BDM$ are alternate angles to lines AC and BD.
This implies,
$AC \parallel\ BD$
$\angle ACB+\angle DBC=180^o$ (Angles on the same side of the transversal are supplementary)
$90^o+\angle DBC=180^o$
$\angle DBC=180^o-90^o$
$\angle DBC=90^o$
Therefore, $\angle DBC$ is a right angle.
(iii) In $\triangle ACB$ and $\triangle DBC$,
$AC=DB$ ($\triangle AMC \cong\ \triangle DMB$, CPCT)
$BC=BC$ (Common side)
$\angle ACB=\angle DBC=90^o$
Therefore,
$\triangle ACB \cong\ \triangle DBC$ (By SAS congruence rule)
(iv) $\triangle ACB \cong\ \triangle DBC$
This implies,
$AB=DC$ (CPCT)
$\frac{1}{2}AB=\frac{1}{2}DC$....(i)
$DM=CM=\frac{1}{2}DC$....(ii)
From (i) and (ii),
$CM=\frac{1}{2}AB$
Hence proved.
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