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
Answers
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
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