Question

In right triangle ABC, right angled 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 (see Fig. 7.23). Show that: 

(i) ∆ AMC ≅ ∆ BMD 

(ii) ∠ DBC is a right angle. 

(iii) ∆ DBC ≅ ∆ ACB (iv) CM = \(\frac{1}{2}\) AB

Answer 1

(i) In \( \triangle AMC \) and \( \triangle BMD \): 

Given that:

\[ AM = BM \quad \text{(M is the mid-point of AB)} \]

\[ \angle AMC = \angle BMD \quad \text{(Vertically opposite angles)} \]

\[ CM = DM \quad \text{(Given)} \]

By the SAS congruence rule:

\[ \triangle AMC \cong \triangle BMD \]

By CPCT (Corresponding Parts of Congruent Triangles):

\[ AC = BD \quad \text{and} \quad \angle ACM = \angle BDM \]

(ii) Since \( \angle ACM = \angle BDM \),

We observe that \( \angle ACM \) and \( \angle BDM \) are alternate interior angles. Since alternate interior angles are equal, it follows that:

\[ DB \parallel AC \]

Now, using the property of co-interior angles:

\[ \angle DBC + \angle ACB = 180^\circ \] \[ \angle DBC + 90^\circ = 180^\circ \] \[ \therefore \angle DBC = 90^\circ \]

(iii) In \( \triangle DBC \) and \( \triangle ACB \):

Since:

\[ DB = AC \quad \text{(Already proved)} \] \[ \angle DBC = \angle ACB = 90^\circ \] \[ BC = CB \quad \text{(Common)} \]

By the SAS congruence rule:

\[ \triangle DBC \cong \triangle ACB \]

(iv) By CPCT:

\[ \angle AB = \angle DC \] \[ \angle AB = 2 \times CM \] \[ \therefore CM = \frac{1}{2} AB \]