Question

in a right triangle,prove that the line segment joining the mid point of the hypotenuse to the opposite vertex is half the hypertenuse

Answer 1

Answer:

Step-by-step explanation:

Here,in triangleABC,<B=90°and D is the midpoint of AC.Produce BD to E such that BD=DE.Join CE.

In triangleADB and CDE ,

AD=CD       (GIVEN)

BD=DE        (BY CONSTRUCTION)

<ADB=<EDC      (VOA)

BY SAS RULE,

triangleADB congruent to triangle CDE

=EC=AB and <CED=<ABD.......i    (by CPCT)

Thus transversal BE cuts AB and EC such that the alternate angles <CEB and<ABE are equal.

AB||EC

<ABC+<ECB=180 (CO INTERIOR)

90+<ECB=180

<ECB=90

Now in triangle ABC and ECB

   AB=EC....(USING (i))

   BC=BC....(common)

   <ABC=<ECB...(EACH 90)

triangle ABC congruent to triangle ECB

      AC=EB.....(CPCT)

      1/ 2AC=1/2EB

       1/2 AC=BD

HOPE THIS HELPS YOU

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Answer 2

Step-by-step explanation:

Given: In ΔABC, ∠B = 90°

D is midpoint of AC

To Prove:

Construction: Draw a line l passing through D and parallel to AB

Proof: Since l || AB

∠ABC = ∠DEC = 90° → (1)

In ΔABC, D is midpoint of AC

DE || AB

∴ E is midpoint of BC (By converse of Midpoint theorem)

⇒ BE = EC → (2)

In ΔDEB and ΔDEC

BE = EC [From (2)]

∠DEB = ∠DEC = 90º [from (1)]

DE is common side

ΔDEB ≅ ΔDEC (By SAS property) ⇒DC = BD

But BD=1/2AC [D is midpoint of AC]

= BD=1/2AC

Hope it will help you