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 #/_CED45^o#
How to show?

Question

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?

Sophia Alfred · Accepted Answer

In acute angled#DeltaBAC#,#/_BAC<90^o# and #/_CADor/_EAD=1/2/_BAC# ,AD being bisector of angle #/_BAC#
So#/_CAD<45^O#
In#DeltaEAO,# #/_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

Wyatt Anderson · Answer

Some thoughts to begin an answer...

OK, here's a diagram to start:

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#

Aaliyah Anderson · Answer

undefined To show that angle $CED$ is greater than $45^\circ$, we can use trigonometry and the properties of triangles.

Since $BE$ is perpendicular to $AC$, triangle $ABC$ is a right triangle. Let's denote the angles of triangle $ABC$ as follows:

Angle $B$ = $\angle BED$

Angle $A$ = $\angle BEA$

Angle $C$ = $\angle CED$

Using the properties of the bisector, we know that $BD/CD = AB/AC$.

By applying the sine rule in triangles $ABE$ and $ACD$, we have:

$\frac{{BD}}{{AB}} = \frac{{\sin A}}{{\sin \angle ABE}}$

and

$\frac{{CD}}{{AC}} = \frac{{\sin A}}{{\sin \angle CAD}}$.

Since $\angle ABE$ and $\angle CAD$ are complementary angles, $\sin \angle ABE = \cos \angle CAD$.

So, $\frac{{BD}}{{AB}} = \frac{{\sin A}}{{\cos \angle CAD}}$ ...(1)

Similarly, from triangle $ACD$, we have:

$\frac{{CD}}{{AC}} = \frac{{\sin A}}{{\cos \angle BAD}}$ ...(2)

From equations (1) and (2), we get:

$\frac{{BD}}{{CD}} = \frac{{\cos \angle CAD}}{{\cos \angle BAD}}$.

By the angle bisector theorem, we know $BD/CD = AB/AC$. Therefore, we can equate the expressions:

$\frac{{AB}}{{AC}} = \frac{{\cos \angle CAD}}{{\cos \angle BAD}}$.

This implies $\frac{{\cos \angle CAD}}{{\cos \angle BAD}} < 1$, since $AB < AC$ in an acute-angled triangle.

Hence, $\cos \angle CAD < \cos \angle BAD$.

As $CAD$ and $BAD$ are acute angles, their cosines are positive. Therefore, $CAD < BAD$, meaning angle $CED$ is greater than angle $BAD$.

Since $BAD$ is an acute angle (as $ABC$ is an acute-angled triangle), $CED$ must be greater than $45^\circ$.

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&gt;45^o# How to show?

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?