ABCD is a triangle right angled at C . A line through the mid point M of hypotenuse AB and parrel to BC intersects AC at D . Show that
1) D is the midpoint of AC
2)MD 1 AC
3) CM = MA = 1/2 AB
Answers
Step-by-step explanation:
Given : A triangle ABC, in which ∠C = 90° and M is the mid-point of AB and BC || DM.
To Prove : (i) D is the mid-point of AC [Given]
(ii) DM ⊥ BC
(iii) CM = MA = 1/ 2 AB
Construction : Join CM.
Proof : (i) In ∆ABC,
M is the mid-point of AB. [Given]
BC || DM [Given]
D is the mid-point of AC [Converse of mid-point theorem] Proved
(ii) ∠ADM = ∠ACB [∵ Coresponding angles]
But ∠ACB = 90° [Given]
∴ ∠ADM = 90°
But ∠ADM + ∠CDM = 180° [Linear pair]
∴ ∠CDM = 90°
Hence, MD ⊥ AC Proved
(iii) AD = DC …(1) [∵ D is the mid-point of AC]
Now, in ∆ADM and ∆CMD, we have
∠ADM = ∠CDM [Each = 90°]
AD = DC [From (1)] DM = DM [Common]
∴ ∆ADM ≅ ∆CMD [SAS congruence]
⇒ CM = MA …(2) [CPCT]
Since M is mid-point of AB,
∴ MA = 1/ 2 AB …(3)
Hence, CM = MA = 1 /2 AB Proved. [From (2) and (3)]
