Question

ABC is a triangle right angled at C. A line through the mid-

point M of hypotenuse AB and parallel to BC intersects ACat D. Show that

(i) D is the mid-point of AC

(ii) MD ⊥ AC

(iii) CM = MA = 1/2 AB

16. VISIT TO FETE

Rahul goes to a fete in Musso​

Answer 1

Answer:

REF.Image

Given : AM=MB & MD∥BC

Since MD∥BC

∠ADM=∠ACB (Corresponding angles)

∠ADM=90

∘

Hence , MD⊥AC

Proved.

Since DM∥BC, By Basic proportionality Theorem

(

AC

AD

)=(

AB

AM

)=

2

1

⇒AC=2AD

Thus D is mid point of AC Proved!

In ΔADM & ΔCDM

AD=CD (proved above)

∠ADM=∠CDM (=90

∘

)

DM=DM (common side)

∴ΔADM≅ΔCDM by SAS

Hence by cpct

AM=CM

But AM=

2

1

AB

∴AM=CM=

2

1

(AB)