Question

HI GUYS HERE'S MY QUESTION

HOW TO PROVE THAT THE MID POINT OF THE HYPOTENUSE OF A RIGHT ANGLED TRIANGLE IS EQUIDISTANT FROM ALL THREE VERTICES

Answer 1

$Answer -$

Construction : Drawn a line parallel to BC from P which cuts AB at D. Join PB.

Given,

PA = PC ------(i)

Let P be the mid point of the hypotenuse of triangle ABC which is right angled at B.

In triangles, PAD and PDB,

PD = PD [common]

angle PDA = angle PDB [converse of mid point theorem]

AD = DB [since D is the point dividing AB into two halves]

Therefore both the triangles are congruent by SAS criteria.

Now, by C.P.C.T

PA = PB -----(ii)

From (i) & (ii),

$PA = PB = PC$

$Proved \: !!..$

Answer 2

In the given ΔABC, D is the midpoint.

We have A at (2a,0), B at (0,2b) and C at (0,0).

The midpoint of (0,2b) and (2a,0) is the midpoint of hypotenuse.

D = (0 + 2a/2, 2b + 0/2)

= > (2a/2, 2b/2)

= > (a,b).

Therefore, it is at (a,b).

Now,

(i)

DB = √(x2 - x1)^2 + (y2 - y1)^2

⇒ √(0 - a)^2 + (2b - b)^2

⇒ √a^2 + b^2

(ii)

DA = √(x2 - x1)^2 + (y2 - y1)^2

⇒ √(2a - a)^2 + (0 - b)^2

⇒ √a^2 + b^2.

(iii)

DC = √(x2 - x1)^2 + (y2 - y1)^2

⇒ √(0 - a)^2 + (0 - b)^2

⇒ √a^2 + b^2.

∴ DB = DA = DC.

Hence, D is equidistant from the ABC.

Hope this helps!