In triangle ABC, AB=AC. Orthocentre lies on Incircle. Find the ratio of AB:BC
Answers
hiiii.......... question not understood
Answer:
Step-by-step explanation:
Let ABC be triangle in which AB = AC. Suppose the orthocentre of the triangle lies on the
in-circle. Find the ratio AB/BC.
Solution: Since the triangle is isosceles, the orthocentre lies on the perpendicular AD from A on
to BC. Let it cut the in-circle at H. Now we are
given that H is the orthocentre of the triangle.
Let AB = AC = b and BC = 2a. Then BD = a.
Observe that b > a since b is the hypotenuse and
a is a leg of a right-angled triangle. Let BH meet
AC in E and CH meet AB in F. By Pythagoras
theorem applied to 4BDH, we get
BH2 = HD2 + BD2 = 4r
2 + a
2
,
where r is the in-radius of ABC. We want to compute BH in another way. Since A, F, H, E are
con-cyclic, we have
BH · BE = BF · BA.
But BF · BA = BD · BC = 2a
2
, since A, F, D, C are con-cyclic. Hence BH2 = 4a
4/BE2
. But
BE2 = 4a
2 − CE2 = 4a
2 − BF2 = 4a
2 −
2a
2
b
2
=
4a
2
(b
2 − a
2
)
b
2
.
This leads to
BH2 =
a
2
b
2
b
2 − a
2
.
Thus we get
a
2
b
2
b
2 − a
2
= a
2 + 4r
2
.
This simplifies to (a
4/(b
2 − a
2
)) = 4r
2
. Now we relate a, b, r in another way using area. We know
that [ABC] = rs, where s is the semi-perimeter of ABC. We have s = (b + b + 2a)/2 = b + a. On
the other hand area can be calculated using Heron’s formula::
[ABC]
2 = s(s − 2a)(s − b)(s − b) = (b + a)(b − a)a
2 = a
2
(b
2 − a
2
).
Hence
r
2 =
[ABC]
2
s
2
=
a
2
(b
2 − a
2
)
(b + a)
2
.
Using this we get
a
4
b
2 − a
2
= 4
a
2
(b
2 − a
2
)
(b + a)
2
.
Therefore a
2 = 4(b − a)
2
, which gives a = 2(b − a) or 2b = 3a. Finally,
AB
BC =
b
2a
=
3
4