ABC is an acute angled triangle. The bisector of /_ BAC intersects BC at D. BE is a perpendicular drawn on AC from B. Points E and D are joined. Show that /_CED>45^o How to show?

2 Answers
Apr 30, 2016

Some thoughts to begin an answer...

Explanation:

OK, here's a diagram to start:

enter image source here

Let:

{ (A = (0, 0)), (B = (e, h)), (C = (c, 0)), (D = (f, k)), (E = (e, 0)), (F = (f, 0)) :}

Then:

sin theta = (DF)/(AD) = k/sqrt(f^2+k^2)

cos theta = (AF)/(AD) = f/sqrt(f^2+k^2)

sin 2theta = (BE)/(AB) = h/sqrt(e^2+h^2)

cos 2theta = (AE)/(AB) = e/sqrt(e^2+h^2)

From the double angle formulae we have:

e/sqrt(e^2+h^2) = cos 2theta = cos^2theta - sin^2theta = (f^2-k^2)/(f^2+k^2)

h/sqrt(e^2+h^2) = sin 2theta = 2cos theta sin theta = (2fk)/(f^2+k^2)

Since D lies on BC there is some t in [0, 1] such that:

(f, k) = (c(1-t)+et, ht)

We want to show that if ABC is acute then DF > EF.

In other words, k > f - e.

Since the triangle is acute, there are some other conditions we can throw in:

e > 0

e < c

AB^2+BC^2 > AC^2, i.e. sqrt(e^2+h^2)+sqrt((c-e)^2+h^2) > c^2

May 1, 2016

enter image source here

In acute angledDeltaBAC,/_BAC<90^o and /_CADor/_EAD=1/2/_BAC ,AD being bisector of angle /_BAC
So/_CAD<45^O
InDeltaEAO, /_AEO=90^o,/_EAO<45^o,hence/_AOE>45^o
:. AE>OE
EP is taken equal in length of OE then P,O are joined and produced.
E,D are also joined and produced ,which meets the produced PO at Q.
Now in DeltaPOE,PE=OE and /_PEO=90^o
So /_EPO=/_EOP=45^o

Now in DeltaPQE,
the exterior/_CEQ=/_CED=/_EPQ+/_PQE
=>/_CED=45^o +/_PQE "Since"/_EPQ=/_EPO=45^o

Hence /_CED>45^o Proved