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 figure). Show that
(i) AAMC = ABMD
(ii) ZDBC is a right angle.
(ii) ADBC = AACB
(iv) CM = 2 AB -----[3]
Answers
Step-by-step explanation:
In Congruent Triangles corresponding parts
are always equal and we write it in short CPCT i e, corresponding parts of Congruent
Triangles.
It is necessary to write a correspondence
of vertices correctly for writing the congruence of triangles in symbolic form.
Criteria for congruence of triangles:
There are 4 criteria for congruence of
triangles.
SAS( side angle side):
Two Triangles are congruent if two sides
and the included angle of a triangle are equal to the two sides and included
angle of the the other triangle.
----------------------------------------------------------------------------------------------------
Given:
In
right angled ∆ABC,
∠C = 90°,
M is the
mid-point of AB i.e, AM=MB & DM = CM.
To Prove:
i) ΔAMC ≅ ΔBMD
ii) ∠DBC is a right angle.
Proof:
(i) In ΔAMC & ΔBMD,
AM = BM (M is the mid-point)
∠CMA = ∠DMB (Vertically opposite angles)
CM = DM (Given)
Hence, ΔAMC ≅ ΔBMD
( by SAS
congruence rule)
ii) since, ΔAMC ≅ ΔBMD
AC=DB.
(by CPCT)
∠ACM = ∠BDM (by CPCT)
Hence, AC || BD as alternate interior angles are equal.
Then,
∠ACB + ∠DBC = 180° (co-interiors angles)
⇒
90° + ∠B = 180°
⇒ ∠DBC = 90°
Hence, ∠DBC = 90°
(ii) In ΔDBC & ΔACB,
BC = CB (Common)
∠ACB = ∠DBC (Right angles)
DB = AC ( proved in part ii)
Hence, ΔDBC ≅ ΔACB (by
SAS congruence rule)
(iii) DC = AB (ΔDBC ≅ ΔACB)
⇒ DM + CM =AB
[CD=CM+DM]
⇒ CM + CM
= AB
[CM= DM
(given)]
⇒ 2CM = AB
Hence,
CM=1/2AB
=========================================================
hope this helps ✌️✌️✌️
Step-by-step explanation:
∠AMCand∠BMD
AM=BM
MC=MD
∠AMC=∠BMD(VOA)
SAS criteria
∠AMC≅∠BMD
∠CAM=∠DBM
If opposite interiar angles are equal
AC||BD
AC||BD cut by BC
∠DBC+∠ACB=1800
∠DBC=900
In∠DBCand∠ACB
By SAS criteria
`/_DBC cong /_ACB
DC=AB
CM=1/2CD=1/2AB.