ABC is a triangle right angled at C. A line through the mid point M of the hypotenuse AB and parallel to BC, intersects AC at D. Show that,
(i) D is the mid-point of AC.
(ii) MD⊥AC
(iii) CM=MA=12AB
Open in App
Solution
(i) In △BAC,
M is the mid-point of AB and MD || BC.
D is the mid-point of AC.
(Converse of mid-point theorem) ∴AD=CD
(ii) ∠ACB=∠ADM (Corresponding angles)
Also, ∠ACB=90° ⇒ADM=90° and MD⊥AC
(iii) Consider △AMD and △CMD DM=DM(Common side) ∠ADM=∠CDM (Right angle) AD=CD (D is the mid-point of side AC)
So, by SAS congruence criterion, △AMD≅△CMD ∴CM=MA=12AB