Question

prove that the line segment joining the mid-points of the hypotenuse of a right-angled triangle to its opposite
vertex is half of the hypotenuse.​

Answer 1

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

PLEASE MARK IT AS A BRAINLIEST ANSWER

Answer 2

Answer:

Step-by-step explanation:

Let P be the mid point of the hypotenuse of the right △ABC right angled at B

Draw a line parallel to BC from P meeting B at O

Join PB

In △PAD and △PBD

∠PDA=∠PDB=90

∘

each due to conv of mid point theorem

PD=PD  (common)

AD=DB  (As D is mid point of AB)

So △ PAD and PBD are congruent by SAS rule

PA=PB  (C.P.C.T)

As PA=PC  (Given as P is mid-point)

∴PA=PC=PB