Question

In right triangle ABC, right angled at C, M is the mid -point of hypotenuse AB. C joined M and produced a point D such that \[DM = CM\]. Point D is joined to point B. Show that:
(i) \[\Delta AMC \cong BMD\]
(ii) \[\angle DBC\] is a right angle.
(iii) \[\Delta DBC \cong \Delta ACB\]

Answer 1

Hint: Here we will use the criterions of congruence to prove the given triangles to be congruent and as M is the midpoint of AB this implies that \[AM = MB\] and we will prove the \[\angle DBC\] is a right angle with the help of CPCT.

Complete step-by-step answer:
It is given that,
\[DM = CM\] and M is the midpoint of AB.
This implies, \[AM = MB\]
(i) In \[\Delta AMC\]and \[\Delta BMD\]
\[AM = MB\] (M is the midpoint of AB)
\[\angle AMC = \angle BMD\](vertically opposite angles)
\[DM = CM\](given)
Therefore, \[\Delta AMC \cong \Delta BMD\] by SAS congruence criterion.
\[AC = BD\](by corresponding parts of congruent triangles)………………………………..(1)
(ii) Now since, \[\Delta AMC \cong \Delta BMD\]
Therefore, \[\angle ACM = \angle BDM\](by corresponding parts of congruent triangles)
Now since, \[\angle ACM\] and \[\angle BDM\] are alternate interior angles of two intersecting lines AC and BD.
Now we know that, if a transversal intersects two lines such that any two alternate interior angles are equal then the liens are parallel to each other.
Therefore, applying this property we get:-
\[BD||AC\]
Now considering BC as a transversal of parallel lines BD and AC.
We know that the sum of interior angles on the same side of the transversal are supplementary.
Therefore, applying this property we get:-
\[\angle DBC + \angle ACB = {180^ \circ }\]
\[\angle ACB = {90^ \circ }\](given)
Putting this value in above equation we get:-
\[\angle DBC + {90^ \circ } = {180^ \circ }\]
Solving it further we get:-
\[\angle DBC = {90^ \circ }\]
Hence proved \[\angle DBC\] is a right angle.
(iii) In \[\Delta DBC\] and \[\Delta ACB\]
\[AC = BD\] (by equation 1)
\[\angle DBC = \angle ACB = {90^ \circ }\]
\[BC = CB\](common)
Therefore, \[\Delta DBC \cong \Delta ACB\] by SAS congruence criterion.

Note: Students should note that two triangles can be congruent by the following criterions.
SSS criterion: - In this criterion, each of the corresponding sides of both the triangles are equal.
SAS criterion: - In this criterion, two corresponding sides of each of the triangles and the corresponding angle between the sides are equal.
ASA criterion: - In this criterion, two corresponding angles of each of the triangles and the corresponding side between the angles are equal.
AAA criterion: - In this criterion, each of the corresponding angles of both the triangles are equal.