hi, i have 2 hard geometry questions that i just cannot do. any help would be great

1) In a pentagon ABCDE, AB=AE, BC=ED, and BD=EC Prove that AD=AC

2)In a quadrilateral ABCD, the diagonal AC bisects <BAD and AB=BC=CD. Prove that the other diagonal BD, bisects <ADC

thanx

2. 1. You have to show that m<ADC = m < ACD ( ADC is an isosceles triangle)

3. Originally Posted by shosho
hi, i have 2 hard geometry questions that i just cannot do. any help would be great

1) In a pentagon ABCDE, AB=AE, BC=ED, and BD=EC Prove that AD=AC

2)In a quadrilateral ABCD, the diagonal AC bisects <BAD and AB=BC=CD. Prove that the other diagonal BD, bisects <ADC

thanx
Nice problems. Good for practice.
Let me show it not in statements-reasons form.

1) In a pentagon ABCDE, AB=AE, BC=ED, and BD=EC Prove that AD=AC
You have to sketch the figure on paper. I do not know how to sketch figures on the board.

The idea is to show that triangle ABC is congruent to riangle AED so that AD=AC.

First, triangle BCD is congruent to triangle EDC by SSS, because
BC=ED, BD=EC as givens, and CD=DC for the one and the same side.
So, angle CBD = angle DEC. -------------(i)

Draw line segment BE.
Triangles CBE and DEB are conguent, by SSS again.
BC=ED, BD=EC as givens, and BE=BE for the one and the same side.
So, angle DBE = angle CEB. -------------(ii)

In triangle ABE, since AB=AE, then the triangle is isosceles, and so,
angle EBA = angle BEA ------------------(iii)

So, using (i),(ii) and (iii), anlge ABC = angle AED.
Hence, triangles ABC and AED are congruent, by SAS, because
AB=AE, angle ABC = angle AED, and BC=ED.

==================================
2)In a quadrilateral ABCD, the diagonal AC bisects <BAD and AB=BC=CD. Prove that the other diagonal BD, bisects <ADC

This one is tougher. Took me many tries. The SSA (or ASS) was a block.

Draw the figure on paper. Draw the diagonal AC.

AB=BC, so triangle ABCis isosceles, and so angle BAC = angle BCA.

angle BAC = angle DAC -------given, or they are half each of angle BAD.

So angle DAC = angle BAC = angle BCA.
Hence, side AD is parallel to side BC. -----reverse of alternate interior angles of parallel lines (and a transversal line, diagonal AC here) are equal.

So, using diagonal BD as the transversal,
angle ADB = angle CBD ----alternate interior angles are equal. ----(a)

But in isosceles triangle BCD, since BC=CD,
angle CBD = angle CDB -----------------------(b)

So, using (a) and (b),

Therefore, diagonal BD bisects angle ADC.

4. Originally Posted by tukeywilliams
1. You have to show that m<ADC = m < ACD ( ADC is an isosceles triangle)
Construct the perpendicular bisector of $\displaystyle CD$ this passes through
the point of intersection of $\displaystyle EC$ and $\displaystyle BD$ it also passes through $\displaystyle A$.
(sorry this does not quite look right in the attachment, but the extra line
is the perpendicular bisector that I require, can you see why it passes through
$\displaystyle A$ and the point of intersection of $\displaystyle EC$ and $\displaystyle BD$?)

The triangles formed by the bisector the two halves of $\displaystyle CD$ and $\displaystyle AC$
and $\displaystyle AD$ are right triangles with two equal sides and so congruent,
and as $\displaystyle AC$ and $\displaystyle AD$ are corresponding sides in these triangles they
are congrent.

RonL

but captainblack.. i like how your method is short and concise but imnot sure how i can prove that the perpendicular goes through the point of intersection?? could you please explain?

6. Originally Posted by shosho

but captainblack.. i like how your method is short and concise but imnot sure how i can prove that the perpendicular goes through the point of intersection?? could you please explain?
consider the bisector of CD that passes through the point of intersection
and label the mid point of CD M and the point of intersection I for
convenience.

Then CI is congruent to DI as triangles DBC and CED are congruent and CD
is a common side (angles BDC and ECD are corresponding angles in these
triangles and so equal which makes triangle DIC isosceles). So triangles CIM
and DIM are congruent by SSA. But angles CMI and DMI are corresponding
angles and so congruent, but they are also supplementary angles and so sum
to 180 degrees, so each or them is 90 degrees.

Hence the line through I and M is a perp bisector of CD.

(alternatively it is obvious by symmetry that the bisector of CD passing
through M is a perp bisector)

RonL

7. Originally Posted by shosho

but captainblack.. i like how your method is short and concise but imnot sure how i can prove that the perpendicular goes through the point of intersection?? could you please explain?
An even shorter demonstration follows from the observation that the
construction has reflection symmetry about MI produced (the bisector of CD through I).

Then as AC is the image of AD under reflection in the sysmmetry line they
have the same length.

RonL

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# ancde is a pentagon inscribed in a circle,,bae= 120

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