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 the given figure). Show that:

1. ΔAMC ≅ ΔBMD
2. ∠DBC is a right angle.
3. ΔDBC ≅ ΔACB
4. CM = `1/2` AB

Answer 1

Since M is the mid-point of AB.

∴ BM = AM

i. In ΔAMC and ΔBMD, we have

CM = DM         ...[Given]

∠AMC = ∠BMD        ...[Vertically opposite angles]

AM = BM                 ...[Proved above]

∴ ΔAMC ≅ ΔBMD       ...[By SAS congruency]

ii. Since ΔAMC ≅ ΔBMD

∠MAC = ∠MBD        ...[By corresponding parts of congruent triangles]

But they form a pair of alternate interior angles.

∴ AC ‖ DB

Now, BC is a transversal which intersects parallel lines AC and DB,

∴ ∠BCA + ∠DBC = 180°      ...[Co-interior angles]

But ∠BCA = 90°            ...[ΔABC is right angled at C]

∴ 90° + ∠DBC = 180°

⇒ ∠DBC = 90°

iii. Again, ΔAMC ≅ ΔBMD           ...[Proved above]

∴ AC = BD                ...[By corresponding parts of congruent triangles]

Now, in ΔDBC and ΔACB, we have

BD = CA              ...[Proved above]

∠DBC = ∠ACB      ...[Each 90°]

BC = CB               ...[Common]

∴ ΔDBC ≅ ΔACB      ...[By SAS congruency]

iv. As ΔDBC ≅ ΔACB

⇒ DC = AB           ...[By corresponding parts of congruent triangles]

But DM = CM       ...[Given]

∴ CM = `1/2` DC = `1/2` AB

⇒ CM = `1/2` AB.