Question

D
A
M
In right triangle ABC, right angled at C, Mis
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. 7.23). Show that:
(i) A AMC = A BMD
ü) DBC is a right angle.
(ii) A DBC = A ACB
B
C
Fig. 7.23
1
(iv) CM =
AB​

Answer 1

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= 1/2 AB

As △DBC≅△ACB

cm= Dm/2

∴DC=AB(△DBC≅△ACB)

So, CM = AB/2

∴CM= 1/2 AB