ABC is a triangle right angled at C. A line through the midpoint M of hypotenuse AB and parallel to BC intersects AC at D.Show that:
(i)D is the midpoint of AC. (ii)MD ,T, AC.
(iii)CM=MA=1\2AB
Please solve these question with proof..
Answers
Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D.
To Prove:
(i) D is the mid - point of AC
(ii) MD | AC
(iii) CM = MA = 1 / 2 AB.
[tex]\texbf{ Proof : (i) Since M is the mid point of hyp. AB and MD | | BC . }[/tex]
⇒ D is the mid - point of AC .
(ii) Since ∠BCA = 90°
and MD | | BC [given]
⇒ ∠MDA = ∠BCA
= 90° [corresp ∠s]
⇒ MD | AC
(iii) Now, in △ADM and △CDM
MD = MD [common]
∠MDA = ∠MDC [each = 90°]
AD = CD [∵ D the mid - point of AC]
⇒ △ADM ≅ △CDM [by SAS congruence axiom]
⇒ AM = CM
Also, M is the mid - point of AB [given]
⇒ CM = MA = 1 / 2 = AB.
this is ur full explanation
PLZ mark as brainliast answer
⛦Given:
- A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D.
⛦To Prove:
(i) D is the mid - point of AC
(ii) MD | AC
(iii) CM = MA = 1 / 2 AB.
⛦Proof :
(i) Since M is the mid point of hyp. AB and MD | | BC .
⇒ D is the mid - point of AC .
(ii) Since ∠BCA = 90°
and MD | | BC [given]
⇒ ∠MDA = ∠BCA
= 90° [corresp ∠s]
⇒ MD | AC
(iii) Now, in △ADM and △CDM
MD = MD [common]
∠MDA = ∠MDC [each = 90°]
AD = CD [∵ D the mid - point of AC]
⇒△ADM ≅ △CDM
[by SAS congruence axiom]
⇒ AM = CM
Also, M is the mid - point of AB [given]
⇒ CM = MA = 1 / 2 = AB
