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.). Show that:
(i) ΔAMC ΔBMD
(ii) DBC is a right angle.
(iii) ΔDBC ΔACB
(iv) CM = \(\frac{1}{2}\) AB
It is given that M is the mid-point of the line segment AB, C = 90°, and DM = CM
(i) Consider the triangles ΔAMC and ΔBMD:
AM = BM (Since M is the mid-point)
CM = DM (Given in the question)
CMA = DMB (They are vertically opposite angles)
So, by SAS congruency criterion, ΔAMC ΔBMD.
(ii) ACM = BDM (by CPCT)
∴ AC BD as alternate interior angles are equal.
Now, ACB +DBC = 180° (Since they are co-interiors angles)
⇒ 90° + B = 180°
∴ DBC = 90°
(iii) In ΔDBC and ΔACB,
BC = CB (Common side)
ACB = DBC (They are right angles)
DB = AC (by CPCT)
So, ΔDBC ΔACB by SAS congruency.
(iv) DC = AB (Since ΔDBC ΔACB)
⇒ DM = CM = AM = BM (Since M the is mid-point)
DM + CM = BM+AM
Hence, CM + CM = AB
⇒ CM = (\(\frac{1}{2}\)) AB