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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.

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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 $BH\perp CP$.

Figure

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    $\begingroup$ Thanks for detailed explanation + thumbs up for geogebra :) $\endgroup$ – markich Jan 5 '14 at 21:51
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I hope the following can satisfy the requirement of a geometrical solution.

I modified Ragnar’s drawing as follow

As pointed out:-

(1) HD = (1/2) AD ------------------------ [intercept theorem]

(2) △ABD ∼ △MCH-------------------- [equi-angular]

Then, from (2), AD / BD = MH / CH ----------- (*)

In △HCP, tan y = HP / CH

Then, tan y = (1/2)(MH) / CH -------------------- [given]

In △HBD, tan z = HD / BD

Then tan z = (1/2)(AD) / BD --------------------- [from (1)]

From (*), we have tan y = tan z

Hence, y = z [since none of the angles are more than π.]

Then, C, D, K and B are con-cyclic ---------[converse of angles in the same segment]

angle CKB = angle CDB = 90 degree------ [angles in the same segment]

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  • $\begingroup$ Please take a look on the following question: math.stackexchange.com/questions/2321816/… $\endgroup$ – Blind Jun 15 '17 at 21:15
  • $\begingroup$ @Blind I have seen your question when it was posted. It looks a bit complicated but I will take a second look at it. Just wondering, is that question somehow related to this answer? $\endgroup$ – Mick Jun 16 '17 at 1:35

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