TRIANGLES 147
But AC^2 =AB^2 + BC^2 (Given) (2)
So, AC = PR [From (1) and (2)] (3)
Now, in ABC and PQR,
AB = PQ (By construction)
BC = QR (By construction)
AC = PR [Proved in (3) above]So, ABC ✁ PQR (SSS congruence)
Therefore, ✂ B =✂ Q (CPCT)
But ✂ Q = 90° (By construction)
So, ✂ B = 90° ✄
Note : Also see Appendix 1 for another proof of this theorem.
Let us now take some examples to illustrate the use of these theorems.
Example 10 : In Fig. 6.48, ✂ ACB = 90°
and CD ☎ AB. Prove that
2
2
BC BD
AC AD
✆ ✝
Solution : ACD ~ ABC
(Theorem 6.7)So,
AC
AB
=
AD
AC
or, AC^2 = AB. AD (1)
Similarly, BCD ~ BAC (Theorem 6.7)
So,
BC
BA
=
BD
BC
or, BC^2 = BA. BD (2)
Therefore, from (1) and (2),
2
2BC
AC
=
BA BD BD
AB AD AD
✝ ✆
✝
Example 11 : A ladder is placed against a wall such that its foot is at a distance
of 2.5 m from the wall and its top reaches a window 6 m above the ground. Find the
length of the ladder.
Fig. 6.48