In right triangle ABC, right angles 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 fig.). Show that :
(i) △AMC=△BMD
(II) ∠DBC is a right angle.
(iii) △DBC=△ACB
(iv) CM=1/2 AB
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In △BMB and △DMC ,
We have
(i) DM = CM (given)
BM = AM (M is the midpoint of AB)
∠DMB = ∠AMC (Vertically opposite angles)
△AMC = △BMD (By SAS)
Hence Proved.
(ii) AC || BD (∠DBM and ∠CAM are alternate angles)
⇒∠DBC + ∠ACB=180° (Sum of co-interior angles)
⇒∠DBC = 90°
Hence proved.
(iii) In △DBC amd △ACB, We have
DB = AC (CPCT)
BC = BC (Common)
∠DBC = ∠ACB (Each=90°)
△DBC = △ACB (By SAS)
Hence proved.
(iv) AB = CD
⇒1/2 AB = 1/2 CD (CPCT)
Hence, 1/2AB=CM Proved.
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