Question

D
A
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 EA BMD
M M
B.
(ii) Z DBC is a right angle.
Fig. 7.23
(iii) ADBC = A ACB
1
(iv) CM = AB
2​

Answer 1

Correct option is

A △AMC≅△BMD

B ∠DBC is a right angle.

C △DBC≅△ACB

D CM=2/1 AB

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

DC² = DB² +BC² (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= DC/2

∴DC=AB(△DBC≅△ACB)

So, CM= AB/2

∴CM=1/2 AB

Sorry to late answer