In right triangle ABC, right angle is at C, M is the mid-point of hypo

1.	In right triangle ABC, right angle is 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 figure). Show that :(i) ΔAMC≅ΔBMD(ii) ∠DBC is a right angle(iii) ΔDBC≅ ΔACB(iv) CM=1/2*AB
Answer» Given: In right angled ∆ABC, ∠C = 90°,M is the mid-point of AB i.e, AM=MB & DM = CM. To Prove: i) ΔAMC ≅ ΔBMD ii) ∠DBC is a right angle. (iii) ΔDBC ≅ ΔACB CM=1/2AB Proof:(i) In ΔAMC & ΔBMD, AM = BM [M is the mid-point] ∠CMA = ∠DMB (Vertically opposite angles) CM = DM (Given) Hence, ΔAMC ≅ ΔBMD[ by SAS congruence rule] ii) since, ΔAMC ≅ ΔBMD AC=DB. (by CPCT) ∠ACM = ∠BDM (by CPCT) Hence, AC \|\| BD as alternate interior angles are equal. Then,∠ACB + ∠DBC = 180° (co-interiors angles) 90° + ∠B = 180° ∠DBC = 90° Hence, ∠DBC = 90° (iii) In ΔDBC & ΔACB, BC = CB (Common) ∠ACB = ∠DBC (Right angles) DB = AC ( PROVED in part ii) Hence, ΔDBC ≅ ΔACB (by SAS congruence rule) (iv) DC = AB (ΔDBC ≅ ΔACB) DM + CM =AB [CD=CM+DM] CM + CM = AB [CM= DM (given)] 2CM = AB Hence, CM=1/2AB HOPE THIS WILL HELP YOU...

In right triangle ABC, right angle is 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 figure). Show that :(i) ΔAMC≅ΔBMD(ii) ∠DBC is a right angle(iii) ΔDBC≅ ΔACB(iv) CM=1/2*AB

Answer»

Given: In right angled ∆ABC,

∠C = 90°,M is the mid-point of AB i.e, AM=MB & DM = CM.

To Prove: i) ΔAMC ≅ ΔBMD

ii) ∠DBC is a right angle.

(iii) ΔDBC ≅ ΔACB

CM=1/2AB

Proof:(i) In ΔAMC & ΔBMD,

AM = BM [M is the mid-point]

∠CMA = ∠DMB (Vertically opposite angles)

CM = DM (Given)

Hence, ΔAMC ≅ ΔBMD[ by SAS congruence rule]

ii) since, ΔAMC ≅ ΔBMD

AC=DB. (by CPCT)

∠ACM = ∠BDM (by CPCT)

Hence, AC || BD as alternate interior angles are equal.

Then,∠ACB + ∠DBC = 180° (co-interiors angles)

90° + ∠B = 180°

∠DBC = 90°

Hence, ∠DBC = 90°

(iii) In ΔDBC & ΔACB,

BC = CB (Common)

∠ACB = ∠DBC (Right angles)

DB = AC ( PROVED in part ii)

Hence, ΔDBC ≅ ΔACB (by SAS congruence rule)

(iv) DC = AB (ΔDBC ≅ ΔACB)

DM + CM =AB [CD=CM+DM]

CM + CM = AB [CM= DM (given)]

2CM = AB

Hence, CM=1/2AB

HOPE THIS WILL HELP YOU...