Question

in a traingle PQR,if PQ2=PR2+RQ2 then its orthocenter always lies​

Answer 1

This is converse of Pythagoras theorem

We can prove this contradiction sum q

2

=p

2

+r

2

 in ΔPQR while triangle is not a rightangle

Now consider another triangle ΔABC we construct ΔABC AB=qCB=b and C is a Right angle

By the Pythagorean theorem (AC)

2

=p

2

+r

2

But we know p

2

+r

2

=q

2

 and q=PR

So (AB)

2

=p

2

+r

2

=(SR)

2

Since PQ and AB are length of sides we can take positive square roots

AC=PQ

All the these sides ΔABC are congruent to ΔPQR

So they are congruent by sss theorem