. In aright triangle ABC ,right angled at B ,M is the mid point of the hypotenuse AC . B is joined to
M and produced to a point D such that DM=BM . Point D is joined to point C .Show that
(1.) ∆ AMB ≅ ∆ CMD
(2.) ∠DCB is a right angle.
(3.) ∆ ≅ ∆ ABC
(4.) BM = 1
AC
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Answer:
i) △AMC≅△BMD
Proof: As 'M' is the midpoint
BM=AM
And also it is the mid point of DC then
DM=MC
And AC=DB (same length)
∴Therefore we can say that
∴△AMC≅△BMD
ii) ∠DBC is a right angle
As △DBC is a right angle triangle and
DC
2
=DB
2
+BC
2
(Pythagoras)
So, ∠B=90°
∴∠DBC is 90°
iii) △DBC≅△ACB
As M is the midpoint of AB and DC. So, DM=MC and AB=BM
∴DC=AB (As they are in same length)
And also, AC=DB
and ∠B=∠C=90°
By SAS Axiom
∴△DBC≅△ACB
iv) CM=
2
1
AB
As △DBC≅△ACB
CM=
2
DC
∴DC=AB(△DBC≅△ACB)
So, CM=
2
AB
∴CM=
2
1
AB
solution
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