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Proving two lines are perpendicular to each other

Hi all,

I have spent hours trying to solve this maths problem but unfortunately I just cannot figure out how this can be solved.

The question reads: The figure shows a square ABCD. M and N are points on AB and BC respectively such that BM=BN. Take a point P on MC such that BP is perpendicular to MC. Prove that PD is perpendicular to PN.

Attachment 24166

Re: Proving two lines are perpendicular to each other

Are you allowed to use trigonometric relations?

The triangles BPM and CPB are similar, this allows to find coordinates of the point P. Together with D and N, you can calculate the angle.

Re: Proving two lines are perpendicular to each other

Can you please elaborate further? I fail to see why BPM and CPD are similar triangles, and even if this is the case I don't know how to obtain the coordinate of P.

And how do I work out the angle once I know the coordinate of P?

Please help.

Re: Proving two lines are perpendicular to each other

Both triangles have a 90°-angle and the angle BMC and MCB add up to 90°, therefore the angle BMC is equal to the angle CBP.

Quote:

And how do I work out the angle once I know the coordinate of P?

Scalar product of the corresponding vectors. But I think this answes my question as "no".

Re: Proving two lines are perpendicular to each other

Start by showing that the triangles DCP and NBP are similar.

$\displaystyle \angle DCP=\angle CMB=\angle NBP, $

so

$\displaystyle \angle DCP=\angle NBP...............................(1)$

Let $\displaystyle \frac{AB}{MB}=\frac{CB}{NB}=k,$

then

$\displaystyle \frac{CD}{NB}=k................................... ..........(2)$

In the triangles $\displaystyle CBP$ and $\displaystyle CBM,$ we have

$\displaystyle \frac{CP}{BP}=\frac{CB}{BM}=k,$ (both ratios are the tangents of the equal angles $\displaystyle PBC$ and $\displaystyle PMB$).

$\displaystyle \frac{CP}{BP}=k................................... ..........(3)$

$\displaystyle (1),(2)$ and $\displaystyle (3)$ together show that the triangles DCP and NBP are similar.

From that it follows that the angles $\displaystyle CDP$ and $\displaystyle BNP$ are equal in which case angles$\displaystyle CDP$ and $\displaystyle CNP$ sum to $\displaystyle 180$ degrees.

It then follows that since the interior angles of a quadrilateral sum to $\displaystyle 360$ degrees, that, (from the quadrilateral $\displaystyle DCNP$) that the angles $\displaystyle DCN$ and $\displaystyle DPN$ sum to $\displaystyle 180$, in which case $\displaystyle DPN=90$ degrees.

Re: Proving two lines are perpendicular to each other

Hi Bob,

I am sure what you posted is 100% correct.

I am almost there understanding what you are saying.

Can you please tell me why angle DCN + angle DPN = 180 degrees?

Re: Proving two lines are perpendicular to each other

$\displaystyle \angle CNP= 180-\angle BNP=180-\angle CDP,$

$\displaystyle \therefore\qquad \angle CNP+\angle CDP=180.$

The sum of the interior angles of a quadrilateral is 360, so the sum of the remaining two must also be 180.