Question

ABC is a triangle right angled at C. A line through the mid-point M of 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 = \(\frac{1}{2}\)AB

Answer 1

Data: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. 

To Prove: (i) D is the mid-point of AC. 

(ii) MD ⊥ AC 

(iii) CM = MA = AB. 

Proof: 

(i) In ∆ABC, M is the mid point of AB and MB || BC Mid-point of BC (Data) 

∴ D is the mid-point of AC (Mid-point formula) 

(ii) MD || BC 

∠BCD + ∠MDC = 180° 

(Sum of interior angles) 

90 + ∠MDC = 180° 

∠MDC = 180 – 90 

∠MDC = 90° 

∴ MD ⊥ AC 

(iii) In this fig. CM is joined. 

M is the mid-point of AB.

∴ MA = \(\frac{1}{2}\)AB ……….. (i) 

Now, in ∆AMD and ∆CMD. 

AD = DC (proved) 

∠MDC = ∠MDA = 90° 

MD is common. 

∴∆AMD ≅ ∆CMD (SAS Postulate) 

∴ CM = MA …………… (ii) 

Comparing (i) and (ii), 

CM = MA = \(\frac{1}{2}\)AB.