Question

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.



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°



(ii) In ΔDBC & ΔACB,
BC = CB (Common)
∠ACB = ∠DBC (Right angles)
DB = AC ( proved in part ii)


Hence, ΔDBC ≅ ΔACB (by SAS congruence rule)

(iii) DC = AB (ΔDBC ≅ ΔACB)

⇒ DM + CM =AB

[CD=CM+DM]


⇒ CM + CM = AB

[CM= DM (given)]
⇒ 2CM = AB

Hence, CM=1/2AB

Answer 1

Answer:

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