Thread: Geometry - Triangles - Congruency

1. Geometry - Triangles - Congruency

Triangles - Congruency

Here are 3 questions i cant solve & when asking my teacher, he couldnt solve too

Question 1:
Triangle ABC and Triangle DEF are such
that AB=DE , AC=DF, AM Perpendicular to BC
and DN Perpendicular to EF . Prove that Triangle
ABC is Congurent to Triangle DEF
[In the figure
triangles are not joined. they are 2 triangles
drawn, at side of each other & both triangles
doesnt have any line connected.]

Question 2:
The Image of an object placed at point
A before a plane mirror LM is seen at the point
B by an observer at D. Prove that the image is
as far behined as the object is seen in front of
the mirror[No Figure Given]

Question 3:
in Triangle ABC, The bisectors of Angle B and
Angle C meet at P. Through P, a straight line MPN is
Drawn parallel to BC. Prove that MN = BM + CN
[No figure Given]

Note: Please show figures and stuff.

2. I'm assuming these aren't problems assigned for homework otherwise your teacher would know how to work them.

3. As posteed #1 is not true.
However, if you know that AM=DN then it is true.

4. Originally Posted by Jonboy
I'm assuming these aren't problems assigned for homework otherwise your teacher would know how to work them.
Yes this isnt a type of homework. these questions were given in refference books of my Grade. extra questions etc. I use the book inmy tution & my tution teacher cant solve it. i have summer holidays, so i cant ask the teacher at the moment.
I think there is a way to solve the 1st question.

Wont Angle BAC and Angle be equal since the lines creating them are equal?

5. Hello, shaurya!

I think I've got #3 . . .

Question 3
In Triangle ABC, the bisectors of angle B and angle C meet at P.
Through P, a straight line MPN is drawn parallel to BC.
Prove that MN = BM + CN
Code:
                  A
*
*
*  *
*
*     *
*
*        *
*  *
*     * P   *
M *--------*---- N
*     *     *  *
*  *           *
B *-----------------* C

Since BP is the bisector of angle B: $\angle PBM = \angle PBC$
Since $MN \parallel BC:\;\angle MPB = \angle PBC = \angle PBM$ (alt-int angles)
. . Hence: $\Delta MBP$ is isosceles and $BM = MP.$

Since CP is the bisector of angle C: $\angle PCN = \angle PCB$
Since $MN \parallel BC:\:\angle NPC = \angle PCB = \angle NCP$
. . Hence: $\Delta NPC$ is isosceles and $CN = PN.$

Then we have: . $MN \:=\:MP + PN \:=\:BM + CN$

7. Originally Posted by shaurya
Triangles - Congruency

Here are 3 questions i cant solve & when asking my teacher, he couldnt solve too
...
Question 2:
The Image of an object placed at point
A before a plane mirror LM is seen at the point
B by an observer at D. Prove that the image is
as far behined as the object is seen in front of
the mirror[No Figure Given]
...
Hello,

1. draw a sketch of the situation.
2. I assume that you are familiar with physical laws of reflection.

3. You can prove that triangle LMA is congruent to triangle LMB because of ASA.

4. Thus length of AM is equal to length of BM.