Question

3
In right triangle ABC, right angled 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. 7.23). Show that:
(Δ AMC = Δ BMD
(ii) 2 DBC is a right angle.
(iii) A DBC =AACB
Fig. 723
(iv) CM
AB
TO a. srinidhi​

Answer 1

Step-by-step explanation:

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