Question

Fig. 7.22
8. In right triangle ABC, right angled at C. Mis
the mid-point of hypotenuse ABC 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:
( A AMC = A BMD
(i) 2 DBC is a right angle.
(iii) A DBC = A ACB
Fig. 7.23
(iv) CM =
AB​

Answer 1

given is not in photo .You write given by self

Answer 2

Correct 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 (see Fig. 7.23). Show that:

(i) ΔAMC ≅ ΔBMD

(ii) ∠DBC is a right angle.

(iii) ΔDBC ≅ ΔACB

(iv) CM = 1/2 AB

______________________________________

$\huge\star{\orange{\underline{\underline{\mathfrak{Explanation:}}}}}$

It is given that M is the mid-point of the line segment AB, ∠C = 90°, and DM = CM

(i) Consider the triangles ΔAMC and ΔBMD:

AM = BM (Since M is the mid-point)

CM = DM

∠CMA = ∠DMB (They are vertically opposite angles)

So, by SAS congruency criterion, ΔAMC ≅ ΔBMD.

(ii) ∠ACM = ∠BDM (by CPCT)

∴ AC | | BD as alternate interior angles are equal.

Now, ∠ACB + ∠DBC = 180° (Since they are co-interiors angles)

⇒ 90° + ∠B = 180°

∴ ∠DBC = 90

(iii) In ΔDBC and ΔACB,

BC = CB (Common side)

∠ACB = ∠DBC ( right angles)

DB = AC (by CPCT)

So, ΔDBC ≅ ΔACB by SAS congruency.

(iv) DC = AB (Since ΔDBC ≅ ΔACB)

⇒ DM = CM = AM = BM (Since M the is mid-point)

So, DM + CM = BM + AM

Hence, CM + CM = AB

⇒ CM = (½) AB

$\large\star{\purple{\underline{\mathfrak{Hence,Solved!}}}}$