28. In a triangle, if the square of one side is equal to sum of the squares on the other two sides,
prove that the angle opposite to the first side is a right angle.
Answers
Step-by-step explanation:
Given:- ABCABC is a triangle
{AC}^{2} = {AB}^{2} + {BC}^{2}AC
2
=AB
2
+BC
2
To prove:- \angle{B} = 90°∠B=90°
Construction:- Construct a triangle PQRPQR right angled at QQ such that, PQ = ABPQ=AB and QR = BCQR=BC
Proof:-
In \triangle{PQR}△PQR
{PR}^{2} = {PQ}^{2} + {QR}^{2} \quad \left( \text{By pythagoras theorem} \right)PR
2
=PQ
2
+QR
2
(By pythagoras theorem)
\Rightarrow {PR}^{2} = {AB}^{2} + {BC}^{2}..... \left( 1 \right) \quad \left( \because AB = PQ \text{ and } QR = BC \right)⇒PR
2
=AB
2
+BC
2
.....(1)(∵AB=PQ and QR=BC)
{AC}^{2} = {AB}^{2} + {BC}^{2} ..... \left( 2 \right) \quad \left( \text{Given} \right)AC
2
=AB
2
+BC
2
.....(2)(Given)
From equation \left( 1 \right) \& \left( 2 \right)(1)&(2), we have
{AC}^{2} = {PR}^{2}AC
2
=PR
2
\Rightarrow AC = PR ..... \left( 3 \right)⇒AC=PR.....(3)
Now, in \triangle{ABC}△ABC and \triangle{PQR}△PQR
AB = PQAB=PQ
BC = QRBC=QR
AC = PR \quad \left( \text{From } \left( 3 \right) \right)AC=PR(From (3))
\therefore \triangle{ABC} \cong \triangle{PQR} \quad \left( \text{By SSS congruency} \right)∴△ABC≅△PQR(By SSS congruency)
Therefore, by C.P.C.T.,
\angle{B} = \angle{Q}∠B=∠Q
\because \angle{Q} = 90°∵∠Q=90°
\therefore \angle{B} = 90°∴∠B=90°
Hence proved.