Question

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. Show that:


(i) angle AMC ≈ angle BMD
(ii) Angle DBC is a right angle.
(iii) Angle DBC = Angle ACB
${\bf{(iv) CM = \frac{1}{2} AB}}$

${\red{\bf{Don't \: Spam.}}}$


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Answer 1

Answer:

Step-by-step explanation:

In AABC, ZC = 90°

/DMB ΔΑMC → In ADMB & AAMC DM = MC (Given)

AM = MC (Mid point of AB is M)

ZDMB ZAMC (VOA) (Velocity =

opposite angle)

..\triangle DMB\cong \triangle AMC (Anu

SAS$$ rule

then,

DB AC (By CP CT)

=

As M is the mid point of AB, so MC bisect ZC in the AABC,

ZMCB MCA = 45°

ZMCA = ZMOB (Alternate angles)

ZMDB = 45° = ZCDB

& ZMCB = ZDCB = 45°

in ADBC,

ZCDB+ZDBC + ZDCB = 180° [sum of

angle in A = 180]

45° + 45° + ZDBC = 180°

ZDBC = 90⁰

As M is the mid point

ZC in the AABC,

ZMCB = ZMCA = 45°

ZMCA = ZMOB (Alternate angles)

ZMDB

: 45°

=

ZCDB

& MCB = ZDCB = 45°

in ADBC,

ZCDB + ZDBC + ZDCB = 180° [sum of

angle in A = 180]

45° +45° + Z/DBC = 180°

ZDBC = 90°

.. ZDBC is a right angle

In ADBC & AACB,

BC= BC (common) ZACB = ZDBC

= 90° AC (proved through CP CT) DB .. ADBC ≈ AACB (By SAS rule) =

AB = CD (By CP CT)

MC = 1/2 =(CD) (as DM = MC)

= 1/2

(AB) (taken from put (w))

Hence proved

Answer 2

Answer:

Given :-

• 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.

Show That :-

• (i) ∆AMC ≌ ∆BMD
• (ii) ∠DBC is a right angle
• (iii) ∆DBC ≌ ∆ACB
• (iv) CM = ½AB

Solution :-

✭ ∆AMC ≌ ∆BMD :-

Given :

• ∠ACB = 90°
• M is the mid-point of AB
• BM = AM
• DM = CM

To Prove :

• ∆AMC ≌ ∆BMD

Proof :

➵ ∠AMC = ∠BMD (Vertically Opposite Angles)

In ∆AMC and ∆BMD,

➵ CM = DM

➵ AM = BM

➵ ∠AMC = ∠BMD

∴ ∆AMC ≌ ∆BMD [ SAS Criterion]

✭ ∠DBC is a right angle :-

➸ BM = AM [M is the mid-point]

➸ ∠BDM = ∠MCA [Opposite angles of equal sides are equal]

[If alternative angles are equal, then sides DB and AC are parallel]

Hence, their sum of two co-interiors angles results to 180°

➸ ∠DBC + ∠BCA = 180°

➸ ∠DBC + 90° = 180° [∠BCA = 90°]

➸ ∠DBC = 180° - 90°

➸ ∠DBC = 90°

∴ ∠DBC is a right angle.

✭ ∆DBC ≌ ∆ACB :-

➠ ∆DBC = ∆ACB,

➠ AB = DC [Given]

➠ ∠DBC = ∠ACB [Perpendicular]

➠ BC = CB [Common]

∴ ∆DBC ≌ ∆ACB [SAS Criterion]

✭ CM = ½AB :-

➢ DC = AB [CPCT]

➢ ½DC = ½AB

As DM = CM

[Then, DM = CM = ½DC]

∴ CM = ½AB

Hence Proved !