Question

In a right triangle \[ABC\], right angle at \[C,M\] is the mid-point of hypotenuse \[AB\]. \[C\] is joined to \[M\] and produced to a point \[D\] such that \[DM = CM\]. Point \[D\] is joined to point \[B\]. Which of the following is/are correct?

This question has multiple correct options
A. \[\Delta AMC \cong \Delta BMD\]
B. \[\angle DBC\] is a right angle
C. \[\Delta DBC \cong \Delta ACB\]
D. \[CM = \dfrac{1}{2}AB\]

Answer 1

Hint: In this question first of all draw the figure by using the given that which will give us a clear idea of what we have to find. By using SAS congruence rule shows the triangles AMC and BMD are similar triangles. Then show that \[\angle DBC\] is a right angle by using Pythagoras theorem. Also prove that triangles DBC and ACB are congruent by using SAS congruence rule. So, use this concept to reach the solution of the given problem.

Complete Step-by-Step solution:
Given \[\Delta ABC\] is a right-angled triangle at \[C\].
So, \[\angle ACB = {90^0}\]
\[M\] is the mid-point of \[AB\]
So, \[AM = BM...................................\left( 1 \right)\]
\[DM = CM..........................................\left( 2 \right)\]
By using the given data, the diagram will be as shown in the figure:

Clearly, lines \[CD\& AB\] intersect each other.
So, \[\angle AMC = \angle BMD{\text{ }}\left( {{\text{Verticaly opposite angles}}} \right){\text{ }}............\left( 3 \right)\]
In \[\Delta AMC\] and \[\Delta BMD\]
\[
   \Rightarrow AM = BM{\text{ }}\left( {{\text{From }}\left( 1 \right)} \right) \\
   \Rightarrow \angle AMC = \angle BMD{\text{ }}\left( {{\text{From }}\left( 3 \right)} \right) \\
   \Rightarrow CM = DM{\text{ }}\left( {{\text{From }}\left( 2 \right)} \right) \\
  \therefore \Delta AMC \cong \Delta BMD{\text{ }}\left( {{\text{SAS congruence rule}}} \right) \\
\]
Hence proved that \[\Delta AMC \cong \Delta BMD\].
Thus, option A. \[\Delta AMC \cong \Delta BMD\] is correct.
As \[\Delta DBC\]is a right-angled triangle
By Pythagoras theorem i.e., \[{\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Height}}} \right)^2} + {\left( {{\text{Base}}} \right)^2}\], we have
\[ \Rightarrow D{C^2} = D{B^2} + B{C^2}\]
So, \[\angle B = {90^0}\]
\[\therefore \angle DBC = {90^0}\]
Thus, option B. \[\angle DBC\] is a right angle is correct.
As \[M\] is the mid-point of AB and DC, we have
\[DM = MC{\text{ and }}AB = BM\]
\[\therefore DC = AB{\text{ }}\left( {{\text{As they are of the same length}}} \right)\]
Also, we have
\[
   \Rightarrow AC = DB{\text{ }}\left( {{\text{As they are of same length}}} \right) \\
   \Rightarrow \angle B = {90^0}{\text{ }}\left( {{\text{right angle}}} \right) \\
   \Rightarrow \angle C = {90^0}{\text{ }}\left( {{\text{right angle}}} \right) \\
\]
So, by SAS congruence rule \[\Delta DBC \cong \Delta ACB\]
Thus, option C. \[\Delta DBC \cong \Delta ACB\] is correct.
As \[\Delta DBC \cong \Delta ACB\], we have
\[CM = \dfrac{{DC}}{2}\]
\[\therefore DC = AB{\text{ }}\left( {\because \Delta DBC \cong \Delta ACB} \right)\]
So, we have \[CM = \dfrac{{AB}}{2}\]
\[\therefore CM = \dfrac{1}{2}AB\]
Thus, option D. \[CM = \dfrac{1}{2}AB\] is also correct.
Note: If two sides and the included angle of one triangle are equal to the corresponding sides and angle of another triangle, the triangles are said to be congruent by SAS similarity criterion. The vertically opposite angles are always equal. Pythagoras theorem states that \[{\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Height}}} \right)^2} + {\left( {{\text{Base}}} \right)^2}\].