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Answers
★ Given :
Given that, in a right triangle ABC :
⇒ ∠C = 90°
⇒ M is the midpoint of hypotenuse AB
⇒ DM = CM
★ To prove :
We have to prove :
⇒ ∆AMC ≌ ∆BMD
⇒ ∠DBC is a right angle
⇒ ∆DBC ≌ ∆ACB
⇒ CM = ½ AB
★ Proof :
⑴ ∆AMC ≌ ∆BMD
In ∆AMC and ∆BMD :
⇒ CM = DM [Given]
⇒ ∠AMC = ∠DMB [Vertically opposite angles] ⇒ ⇒ AM = BM [M is the midpoint]
∴ By SAS rule, ∆AMC ≌ ∆BMD.
Hence, proved!
Also :
⇒ ∠ACM = ∠BDM [CPCT]
⇒ AC = BD [CPCT]
⑵ ∠DBC is a right angle
Here, ∠ACM and ∠BDM are two alternate interior angles for the lines BD and AC.
If a transversal intersects two lines such that a pair of alternate interior angles are equal, rhen the lines are said to be parallel to each other.
BD ∥ AC.
Considering BD ∥ AC with BC as the transversal :
⇒ ∠DBC + ∠ACB = 180° [Interior angles]
⇒ ∠DBC + 90° = 180°
⇒ ∠DBC = 90°
Hence, proved!
⑶ ∆DBC ≌ ∆ACB
In ∆DBC and ∆ACB :
⇒ BD = AC [CPCT]
⇒ ∠DBC = ∠ACB [Proved]
⇒ BC = BC [Common]
∴ By SAS rule, ∆DBC ≌ ∆ACB.
Also :
⇒ DC = AB [CPCT]
⑷ CM = ½ AB
⇒ DC = AB [Proved]
⇒ DC = AB
⇒ DM + CM = AB [M is the midpoint]
⇒ CM + CM = AB
⇒ 2 CM = AB
⇒ CM = AB/2
⇒ CM = ½ AB
Hence, proved!