Question

c) In right triangle ABC, right angle at C, M is the mid-point of the 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- (4m)
​

Answer 1

Given:

➽∠ACB=90°

➽AM = BM (as M is the midpoint)

➽DM = CM

To Prove:

(i)△AMC ≅ △BMD

(ii)∠DBC =∠ACB

(iii)△DBC ≅ △ACB

(iv) CM=$\dfrac {1}{2} AB$

Solution:

(i)△AMC ≅ △BMD

In △AMC and △BMD

AM = BM (as M is the midpoint)

∠AMC = ∠BMD (vertically opposite angles)

CM = DM (given)

so, △AMC ≅ △BMD (By SAS)

AC = BD (CPCT)

∠MAC = ∠MBD (by CPCT)

(ii)∠DBC =∠ACB

∠MAC and ∠MBD are also alternate interior angles.

Also,

∠MAC = ∠MBD [proved in (i)]

so, AC || BD (If Alternate angles are equal then lines are parallel)

we can say,

∠ACB + ∠DBC = 180° (Sum of Co-interior angle is always 180°)

∠C and ∠B are co-interior as AC || BD

∠ACB + ∠DBC = 180°

90° + ∠DBC = 180°

∠DBC = 180° - 90°

∠DBC = 90°

Hence proved ∠DBC =∠ACB

(iii)△DBC ≅ △ACB

DB = AC [proved in (i)]

∠DBC =∠ACB (Each 90°)

CB = CB (common)

△DBC ≅ △ACB (SAS)

so, DC = AB (By CPCT)

$\bold {(iv)CM= \dfrac {1}{2} AB }$

CM= $\dfrac {1}{2} DC$(as CM =DM)

Since AB = DC, replace AB by DC

Hence proved, CM= $\dfrac {1}{2}$ AB

Answer 2

(i) In △ AMC and △BMD

AM = BM (M is Mid Point of AB)

∠ AMC = ∠ BMD ( Vertically opposite angle )

CM = DM (GIVEN)

∴ △ AMC @ △BMD (By SAS congruence rule)

∴ AC = BD (CPCT)

And ∠ ACM = ∠ BDM by (CPCT)

(ii) We have ∠ ACM = ∠ BDM

But ∠ ACM and ∠ BDM are alternate interior angle

Since alternate angles are equal.

Hence We can say that DB ll AC

⟹ ∠ DBC + ∠ ACB = 180 Degree (Co- Interior Angles)

∠ DBC + 90 Degree = 180 Degree

⟹ ∠ DBC = 90 Degree

(iii) Now in △ DBC and △ ACB

DB = AC ( Already Proved )

∠ DBC = ∠ ACB each (90 Degree)

BC = CB (Common)

∴ △DBC  ≅ △ACB (SAS Congruence Rule)

(iv) We have △ DBC  ≅ △ACB

∴ AB = (DC by CPCT)

⟹ AB = 2 CM

∴ CM = $\frac{1}{2}$ AB