geometry
We have given isosceles triangle $ABC$ with baseline $AB$. Point $M$ is midpoint of AB. Draw perpendicular line to side $AC$ through $M$ which intersects $AC$ in point $H$. Let $P$ be midpoint of $MH$. Show that lines $BH$ and $CP$ are perpendicular to each other.
I also have "hint":
Construct height $BD$ to side $AC$ and look at the triangles $ABD$ and $MCH$.
Analytically there is no problem but I get stuck in geometrical approach to solution.
Triangles $ABH$ and $ABK$ are right triangles whose hypothenuse is $AB$. So they have the same circumcircle, whose diameter is $AB$. As a consequence, considering that $M$ is the midpoint of the diameter $AB$, we get that $MH=MK$ because they are radii of the same circumference, and then $MHK$ is isosceles.
Without using circles: we can show that, in any right triangle, the median drawn to the hypotenuse is equal to half the hypotenuse. To show it for the right triangle $AHB$, let us draw a line $ME$ starting from the midpoint $M$, parallel to the leg $AH$, and ending to the intersection point $E$ with the other leg $HB$. We know that the angle $AHB$ is right. The angles $MEB$ and $AHB$ are congruent, since they are corresponding angles if we consider the parallel lines $AH$ and $ME$, and the transverse line $HB$. So, the angle $MEB$ is a right angle.
Because the line $ME$ starts from the midpoint $M$ and is parallel to $AH$, it divides the leg $HB$ in two congruent segments $HE$ and $EB$ with equal length. These considerations show that the triangles $MEH$ and $MEB$ are right triangles, have congruent legs $HE$ and $EB$, and a common leg $ME$. This means that these triangles are congruent. We then get that the segments $MH$ and $MB$ are also congruent, since they are corresponding sides (hypothenuses) of these triangles. So we have shown that, in the right triangle $AHB$, the median $MH$ is equal to half the hypotenuse $AB$.
Applying the same procedure to the right triangle $BKA$, we get that $MK$ is congruent with $MA$, so that it also equals half of the hypothenuse $AB$.
We then conclude that $MH=MK$, and so the triangle $MHK$ is isosceles.
First off, the other two answers here are wrong because you can prove the statement using the same flaw if P was outside the triangle. The proof using the same flaw but done in the outside of the triangle is here: [https://www.youtube.com/watch?v=Yajonhixy4g][1]
The video clearly does this proof with P on the outside. If P were on the segments of the triangle, then the proof will still hold because triangle AEP will be congruent to triangle AFP by SAA (shared sides, bisected angles, 90 degree angles). You know that P and D are the same points because the definition of P is where the perpendicular bisector meets the bisector of the opposite angle, and provided the case where P's on BC, it can only possibly be a triangle if P and D are on the same place. D is the bisector of BC, so so is P. EP = FP because we proved that AEP = AFP. BP = CP because P bisects and is between BC. Since EBP and FCP are right triangles and the Hypotenuse and one of the legs are congruent, we have triangle EBP is congruent to triangle FCP. From here, basic elementary algebra can be use to prove AB = AC.
So, what is wrong with this proof? The answer is betweenness. Every single one of these proofs rely on a certain order of points your drawing above should be:
We still have AE=AF, PE=PF, and PB=PC, and it's still true that BE=FC, but AB=AC is not true. This is because F is still between A and C, but E is not between A and B. Since E is not between A and B, SAA does not follow, and neither does SAS for this particular case.
Best Answer
We draw the line $BD$ perpendicular to $AC$ and parallel to $MH$. Because $\angle CBA=\angle CAB=\angle CMH$, we know that $\triangle AMH\sim\triangle ABD\sim\triangle MCH$. Because $|AM|=|MB|$, it follows that $|AH|=|HD|$. Thus, $H$ is the midpoint of $AD$. Now, we see that $BH$ and $CP$ are medians in $\triangle ABD$ and $\triangle MCH$. Also, we can get $\triangle MCH$ by rotating $\triangle ABD$ over $90^\circ$ and translating and scaling it. Because translation and scaling don't change the angle between corresponding lines, we now find that $BH\perp CP$.