Question

a triangle ABC right angled at c a line through the midpoint of the hypotenuse a b and parallel to BC intersect AC at d show that cm=ma=1/2 ab
​

Answer 1

Answer:

R.E.F. Image.

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

Step-by-step explanation:

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you can see the picture that I have attached

Answer 2

QUESTION :

ABC is a triangle right angled at C.A. line through the midpoint M of hypotenuse and parallel to BC intersects AC at D. show that

(i) D is the midpoint of AC

(II) MD⊥AC

(iii) CM=MA= 1/2 AB

ANSWER :

R.E.F. Image.In △ABC .

(1) M is the mid point of AB andMD∥BCD is the mid point of AC

(2) As MD∥BC &AC is transversal∠MDC+∠BCD=180 ∘ ∠MDC+90 ∘ =180 ∘ ⇒∠MDC=90 ∘ ⇒MD⊥MC

(3) In △AMD and △CMDAD=CD∠ADM=∠CDMDM=DM△AMD≅△CMDAM=CM⇒AM= 1/2 AB⇒CM=AM= 1/2 AB

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