Maths

TRIANGLES 147
So, ✁
✂ ✂
✂
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]
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°
BC BD
and CD AB. Prove that 2
☎
AC AD
Solution : ACD ~ ABC
✝ ✆
So,
2
(Theorem 6.7)
AC
AB = AD
AC
or, AC 2 = AB . AD (1)
Fig. 6.48
Similarly, BCD ~ BAC (Theorem 6.7)
BC
So,
BA = BD
BC
or, BC 2 = BA . BD (2)
Therefore, from (1) and (2),
BC
AC
2
2
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.