Question

m is the mid point of the hypitenuse AB of a right triangle ABC. prove that Cm=1/2AB
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Answer 1

Answer:

⏮In ΔAMC and ΔBMD,

AM = BM (M is the mid-point of AB)

∠AMC = ∠BMD (Vertically opposite angles) and

CM = DM (Given)

∴ ΔAMC ≅ ΔBMD (By SAS congruence rule)

∴ AC = BD (By CPCT) And, ∠ACM = ∠BDM (By CPCT)

(ii) ∠ACM = ∠BDM

∠ACM and ∠BDM are alternate interior angles.

Since alternate angles are equal, It can be said that

DB || AC ∠DBC + ∠ACB = 180º (Co-interior angles)

∠DBC + 90º = 180º ∠DBC = 90º

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(iii) In ΔDBC and ΔACB,

DB = AC (Already proved above)

∠DBC = ∠ACB (Each 90 )

BC = CB (Common)

∴ ΔDBC ≅ ΔACB (SAS congruence rule) ❤♥

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(iv) ΔDBC ≅ ΔACB

AB = DC (By CPCT)

AB = 2 CM

∴ CM =1/2

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