Question

ABC is a triangle right angled at c. a line at the mid point M of hypotenuse AB and parallel to BC intersect AC at D. show that
1.Dis the mid point of AC
2. MD perpendicular of AC
3. CM = MA = 1/2 AB​

Answer 1

Answer:

In △ABC,

(1) M is the mid point of AB and

MD∥BC

D is the mid point of AC

(2) As MD∥BC &

AC is transversal

∠MDC+∠BCD=180

∘

∠MDC+90

∘

=180

∘

⇒∠MDC=90

∘

⇒MD⊥AC

(3) In △AMD and △CMD

AD=CD

∠ADM=∠CDM

DM=DM

△AMD≅△CMD

AM=CM⇒AM=

2

1

AB⇒CM=AM=

2

1

AB