# Thread: Help with these two geometry!

1. ## Help with these two geometry!

Help with these two hard problems:

Problem 12
Let M be a point inside the angle A. From point M perpendiculars MP and MQ are dropped to the angle sides. From point A, perpendicular AK is dropped to segment PQ. Prove that angle PAK = angle MAQ.
Problem 13
Let ABC be a triangle and angle B = 60. Angle bisectors AD and CE meet at the point O. Prove that OD = OE.

2. Hello xxravenxx
Originally Posted by xxravenxx
... Problem 12
Let M be a point inside the angle A. From point M perpendiculars MP and MQ are dropped to the angle sides. From point A, perpendicular AK is dropped to segment PQ. Prove that angle PAK = angle MAQ.
$\displaystyle \angle APM = \angle AQM = 90^o$ (given)

$\displaystyle \Rightarrow APMQ$ is a cyclic quadrilateral (opp angles supplementary)

$\displaystyle \Rightarrow \angle APQ = \angle AMQ$ (angles in same segment)

So in $\displaystyle \triangle$'s $\displaystyle APK, AMQ:$
$\displaystyle \angle APK = \angle AMQ$ (proven)
$\displaystyle \angle AKP = \angle AQM$ ($\displaystyle =90^o$, given)
$\displaystyle \Rightarrow \angle PAK = \angle MAQ$ (angle sum of triangle)

Grandad

3. Sorry but can you draw me a diagram of what this looks like? I'm a bit clueless.

4. Hello xxravenxx
Originally Posted by xxravenxx
Sorry but can you draw me a diagram of what this looks like? I'm a bit clueless.
See attachment.

... and here's the answer for number 13:
Originally Posted by xxravenxx
...Problem 13
Let ABC be a triangle and angle B = 60. Angle bisectors AD and CE meet at the point O. Prove that OD = OE.
(I hope you can draw a diagram for this.)

Let $\displaystyle \angle BAD = \angle DAC = x^o$ and $\displaystyle \angle ACE = \angle ECB = y^o$

Then, $\displaystyle 2x+2y+ 60=180$ (angle sum of $\displaystyle \triangle ABC$)

$\displaystyle \Rightarrow x+y=60$

Now $\displaystyle \angle DOC$ is an exterior angle of $\displaystyle \triangle OAC$ = sum of interior opposite angles $\displaystyle = x+y = 60^o$

$\displaystyle \Rightarrow BEOD$ is a cyclic quadrilateral (exterior angle $\displaystyle \angle DOC$ = interior opposite angle $\displaystyle \angle B$)

Now the angle bisectors of a triangle are concurrent $\displaystyle \Rightarrow BO$ is the bisector of $\displaystyle \angle ABC$

$\displaystyle \Rightarrow \angle DBO = \angle EBO$

$\displaystyle \Rightarrow \angle DEO = \angle EDO$ (angles in same segment)

$\displaystyle \Rightarrow OE = OD$ (isosceles $\displaystyle \triangle OED$)

Grandad